(a) The area of a rectangular garden is 7371 m².
If the width of the garden is 81 m, what is its length?
What is the Length of the garden:

..........................

Answer:

d

Step-by-step explanation:

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Answer:

Consecutive Interior

Step-by-step explanation:

When two lines are crossed by another line (called the Transversal) The pairs of angles on one side of the transversal but inside the two lines are called Consecutive Interior Angles.

Answer:

$24

Step-by-step explanation:

$24

30 x .80 equals 24

Hope this helps!

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Two triangles that are congruent using SAS.

SAS = Side Angle Side

The two triangles in option B are congruent by SAS.

Option B is the correct answer

It is a two-dimensional figure which has three sides and the sum of the three angles is equal to 180 degrees.

We have,

Two triangles that are congruent using SAS.

SAS = Side Angle Side

Now,

The two triangles in option B are congruent by SAS.

There are two sides that are equal in each triangle.

There is one angle that is equal in each triangle.

Thus,

The two triangles in option B are congruent by SAS.

Option B is the correct answer

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Answer:

(3n - 1)(n + 2)

Step-by-step explanation:

Answer:

la respuesta es 16vpor que aumenta 2

Answer: 6 inches long and 4 inches wide

Step-by-step explanation: The dimensions of the actual crate are 20 times those on Jacob's drawing. The dimensions on the builder's drawing are 1/10 of those, so (1/10)(20) = 2 times the dimensions on Jacob's drawing.

hope this helps

Answer:

Step-by-step explanation:

10% of 26.2 miles = 0.10×26.2 miles = 2.61 miles

Answer:

1 and 2 adjacent

5 and 6 are linear pair

Step-by-step explanation:

Answer:

1/10

Step-by-step explanation:

(4/40)/ (4/4) = 1/10

Answer:

C

Step-by-step explanation:

The possible value of the machine at the end of each year will be $3200

Depreciation in economics is a measure of the amount of value an asset loses from influential factors affecting its market value.

Given that, the value of a machine is decreasing by a constant percent rate each year, we need to find the possible value of the machine at the end of each year,

So, we can say that this is a case of depreciation,

Since, the values are not given, let us consider the rate be 20% and the initial price of the machine be $16000.

We know that, the depreciation, formula is given by :

A = P(1-r)ⁿ

Where, A = final amount, P = initial amount, r = rate and n = time.

Since, the value is decreasing by a constant percent rate each year,

Therefore, every year the amount decreased will be same,

Finding for 1 year,

A = 16000(1-0.2)¹

A = 16000(0.8)

A = 12800

The amount decreased = 16000-12800 = $3200

Therefore, after each year the decreased amount will be $3200

Hence, the possible value of the machine at the end of each year will be $3200

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Answer:

for the number of rubas, it keeps on doubling up while on the square number its only going up by 1 every time.

Step-by-step explanation:

Answer:

I'm pretty sure it's D

Step-by-step explanation:

maths

Answer:

Step-by-step explanation:

15a²bc = 3·5·a²·b·c

20ab²c = 2²·5·a·b²·c

25bc³ = 5²·b·c³

greatest common factor: 5bc

(-15a²bc + 20ab²c - 25bc³)÷(5bc) = -3a² + 4ab - 5c²

-15a²bc + 20ab²c - 25bc³ = (-3a² + 4ab - 5c²)(5bc)

Answer:

a) The boundaries are [tex]Z = \pm 0.253[/tex]

b) The boundaries are [tex]Z = \pm 0.675[/tex].

c) The boundaries are [tex]Z = \pm 1.96[/tex].

d) The boundaries are [tex]Z = \pm 2.575[/tex].

Step-by-step explanation:

Z-score:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

a. The middle 20% from the 80% in the tails.

The middle 20% is between the 50 - (20/2) = 40th percentile and the 50 + (20/2) = 60th percentile:

40th percentile: Z has a pvalue of 0.4, so Z = -0.253.

60th percentile: Z has a pvalue of 0.6, so Z = 0.253.

The boundaries are [tex]Z = \pm 0.253[/tex].

b. The middle 50% from the 50% in the tails.

The middle 50% is between the 50 - (50/2) = 25th percentile and the 50 + (50/2) = 75th percentile:

25th percentile: Z has a pvalue of 0.25, so Z = -0.675.

75th percentile: Z has a pvalue of 0.75, so Z = 0.675.

The boundaries are [tex]Z = \pm 0.675[/tex].

c. The middle 95% from the 5% in the tails.

The middle 95% is between the 50 - (95/2) = 2.5th percentile and the 50 + (95/2) = 97.5th percentile:

2.5th percentile: Z has a pvalue of 0.025, so Z = -1.96.

97.5th percentile: Z has a pvalue of 0.975, so Z = 1.96.

The boundaries are [tex]Z = \pm 1.96[/tex].

d. The middle 99% from the 1% in the tails.

The middle 99% is between the 50 - (99/2) = 0.5th percentile and the 50 + (99/2) = 99.5th percentile:

0.5th percentile: Z has a pvalue of 0.005, so Z = -2.575.

99.5th percentile: Z has a pvalue of 0.995, so Z = 2.575.

The boundaries are [tex]Z = \pm 2.575[/tex].

A z-core is also referred to as a standard score and it can be defined as a measure of the distance between a data point (raw score) and the mean, when standard deviation units are used.

In Statistics, z-scores can either be positive or negative and it is calculated by using this formula:

[tex]Z=\frac{x\;-\;u}{\delta}[/tex]

Where:

a. A middle 20% from the 80% in the tails.

Note: The middle 20% is between the 40th percentile and 60th percentile.

Z = [tex]\pm 0.253[/tex]

b. A middle 50% from the 50% in the tails.

Note: The middle 50% is between the 25th percentile and 75th percentile.

Z = [tex]\pm 0.675[/tex]

c. A middle 95% from the 5% in the tails.

Note: The middle 95% is between the 2.5th percentile and 97.5th percentile.

Z = [tex]\pm 1.96[/tex]

d. A middle 99% from the 1% in the tails.

Note: The middle 99% is between the 0.5th percentile and 99.5th percentile.

Z = [tex]\pm 2.575[/tex]

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Answer:

Your answer is

60% apply means 30k *0.6= 18000

30% enroll so,.0.3*18k = 5400

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Answer:

5400 students

Step-by-step explanation:

well 60% of 30000 would be 18000 which is the amount who gets accepted. 30% of 18000 would be who actually enrolls

Answer:

Step-by-step explanation:

[tex]L=4W-2\\ \\ P=2(L+W)\\ \\ 2(4W-2+W)=36\\ \\ 5W-2=18\\ \\ 5W=20\\ \\ W=4,\ L=14\\ \\ A=4(14)=56cm^2[/tex]

Answer:

L=4W-2

P=2L+2W

36= 2(4W-2)+2W

36= 8W-4 +2W

36= 10W-4

36+4=10W

40=10W

40/10=W

W=4

L=4(4)-2

L=14

AREA =L(W)

AREA = 4X14

AREA =48 cm

Answer:

rupees 14.69

Step-by-step explanation:

rupees 14.69

I hope this answer correct

Answer:

i think it is a but i am not shore. Hope this helps

Step-by-step explanation:

Answer:

PROBLEM 5: 80

PROBLEM 6: 94

Step-by-step explanation:

PROBLEM 5:

5(2 times 5 + 6)

=5(10+6)

5(16) = 5 times 16

=80

PROBLEM 6:

4 times(5 to the power 2)-6

4(25)-6

=100-6

=94

Answer:

[tex]P_{20} = 20[/tex] --- 20th percentile

[tex]P_{25} = 21.75[/tex]  --- 25th percentile

[tex]P_{65} = 27.85[/tex]   --- 65th percentile

[tex]P_{75} = 29.5[/tex]   --- 75th percentile

Step-by-step explanation:

Given

27, 24, 21, 16, 30, 33, 28, and 24.

N = 8

First, arrange the data in ascending order:

Arranged data: 16, 21, 24, 24, 27, 28, 30, 33

Solving (a): The 20th percentile

This is calculated as:

[tex]P_{20} = 20 * \frac{N +1}{100}[/tex]

[tex]P_{20} = 20 * \frac{8 +1}{100}[/tex]

[tex]P_{20} = 20 * \frac{9}{100}[/tex]

[tex]P_{20} = \frac{20 * 9}{100}[/tex]

[tex]P_{20} = \frac{180}{100}[/tex]

[tex]P_{20} = 1.8th\ item[/tex]

This is then calculated as:

[tex]P_{20} = 1st\ Item +0.8(2nd\ Item - 1st\ Item)[/tex]

[tex]P_{20} = 16 + 0.8*(21 - 16)[/tex]

[tex]P_{20} = 16 + 0.8*5[/tex]

[tex]P_{20} = 16 + 4[/tex]

[tex]P_{20} = 20[/tex]

Solving (b): The 25th percentile

This is calculated as:

[tex]P_{25} = 25 * \frac{N +1}{100}[/tex]

[tex]P_{25} = 25 * \frac{8 +1}{100}[/tex]

[tex]P_{25} = 25 * \frac{9}{100}[/tex]

[tex]P_{25} = \frac{25 * 9}{100}[/tex]

[tex]P_{25} = \frac{225}{100}[/tex]

[tex]P_{25} = 2.25\ th[/tex]

This is then calculated as:

[tex]P_{25} = 2nd\ item + 0.25(3rd\ item-2nd\ item)[/tex]

[tex]P_{25} = 21 + 0.25(24-21)[/tex]

[tex]P_{25} = 21 + 0.25(3)[/tex]

[tex]P_{25} = 21 + 0.75[/tex]

[tex]P_{25} = 21.75[/tex]

Solving (c): The 65th percentile

This is calculated as:

[tex]P_{65} = 65 * \frac{N +1}{100}[/tex]

[tex]P_{65} = 65 * \frac{8 +1}{100}[/tex]

[tex]P_{65} = 65 * \frac{9}{100}[/tex]

[tex]P_{65} = \frac{65 * 9}{100}[/tex]

[tex]P_{65} = \frac{585}{100}[/tex]

[tex]P_{65} = 5.85\th[/tex]

This is then calculated as:

[tex]P_{65} = 5th + 0.85(6th - 5th)[/tex]

[tex]P_{65} = 27 + 0.85(28 - 27)[/tex]

[tex]P_{65} = 27 + 0.85(1)[/tex]

[tex]P_{65} = 27 + 0.85[/tex]

[tex]P_{65} = 27.85[/tex]

Solving (d): The 75th percentile

This is calculated as:

[tex]P_{75} = 75 * \frac{N +1}{100}[/tex]

[tex]P_{75} = 75 * \frac{8 +1}{100}[/tex]

[tex]P_{75} = 75 * \frac{9}{100}[/tex]

[tex]P_{75} = \frac{75 * 9}{100}[/tex]

[tex]P_{75} = \frac{675}{100}[/tex]

[tex]P_{75} = 6.75th[/tex]

This is then calculated as:

[tex]P_{75} = 6th + 0.75(7th - 6th)[/tex]

[tex]P_{75} = 28 + 0.75(30- 28)[/tex]

[tex]P_{75} = 28 + 0.75(2)[/tex]

[tex]P_{75} = 28 + 1.5[/tex]

[tex]P_{75} = 29.5[/tex]

Answer:

(0, -2)

Step-by-step explanation:

Crosses the y-axis = find the y-intercept. Substitute x = 0 (if you draw a graph, you can see that the point where y crosses the y axis is always 0 for x) into the equation, and you get 3(0) - 7y = 14. Calculate, you get y = -2. Rewrite in coordinates form.

Answer:

$40

Step-by-step explanation:

400x0.10=40

Answer:

mariana bedoya sanchez

Answer:

27

Step-by-step explanation:

it was right on khan academy

Answer:

The exponential smoothing forecast for period 13 = 39.68

Step-by-step explanation:

As given,

[tex]\alpha[/tex] = 0.4

Time                 Sales                              forcast

1                            48                                  

2                           41                                 48

3                           37                                 48 + [tex]\alpha {(41 - 48)}[/tex] = 45.2

4                           32                                 45.2 + [tex]\alpha {(37 - 45.2)}[/tex] = 41.92

5                           36                                 41.92 + [tex]\alpha {(32 - 41.92)}[/tex] = 37.95

6                           31                                 37.95 + [tex]\alpha {(36 - 37.95)}[/tex] = 37.17

7                            43                                 37.17 + [tex]\alpha {(31 - 37.17)}[/tex] =  34.70

8                           52                                 34.70 + [tex]\alpha {(43 - 34.70)}[/tex] = 38.02

9                           60                                 38.02 + [tex]\alpha {(52 - 38.02)}[/tex] = 43.61

10                          48                                 43.61 + [tex]\alpha {(60 - 43.61)}[/tex] = 50.17

11                           41.9                                 50.17 + [tex]\alpha {(48 - 50.17)}[/tex] = 49.30

12                           29.7                                 49.30 + [tex]\alpha {(41.9 - 49.30)}[/tex] = 46.34

13                                                                    46.34 + [tex]\alpha {(29.7 - 46.34)}[/tex] = 39.68

∴ we get

The exponential smoothing forecast for period 13 = 39.68

Answer:

f(-3) = 4(-3) + 5 = -12 + 5 = -7. the answer is #2.

Answer:

D

8-2

Step-by-step explanation:

System of inequalities are

x + y ≥ 7

8x + 7y ≤ 82