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Lostsunrise [7]

3 years ago

10

A sand box has an area of 45ft. The length is 4 feet longer than the width. Find the dimensions of the sand box. Solve by comple

ting the sqaure

1 answer:

choli [55]3 years ago

3 0

The length is 9ft and the width is 5ft.

You might be interested in

Which are the complementary angles?

Luda [366]

Answer:

Two angles are called complementary when their measures add to 90 degrees. Two angles are called supplementary when their measures add up to 180 degrees.

Step-by-step explanation:

please amrk this answer as brainliest

8 0

3 years ago

(Hypothetical.) Suppose five measurements of the volume of a liquid yield the following data (in milliliters).

Digiron [165]

Answer:

The 95% confidence interval for the measurements is [48.106, 53.494].

Step-by-step explanation:

The average M of this sample is

$M=\dfrac{1}{5}\sum_{i=1}^5x_i=\dfrac{1}{5}(52+48+49+52+53)\\\\\\M=\dfrac{254}{5}=50.800$

The standard deviation s of this sample is

$s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^5(x_i-M)^2}$

$s=\sqrt{\dfrac{1}{4}\cdot [(52-50.8)^2+(48-50.8)^2+(49-50.8)^2+(52-50.8)^2+(53-50.8)^2]}$

$s=\sqrt{\dfrac{1}{4}\cdot [(1.44)+(7.84)+(3.24)+(1.44)+(4.84)]}=\sqrt{\dfrac{18.8}{4}}=\sqrt{4.7}\\\\\\s=2.168$

The degrees of freedom are

$df=n-1=5-1=4$

Then, the critical value of t for a 95% CI and 4 degrees of freedom is t=2.776.

The margin of error of the CI is:

$E=t\cdot s/\sqrt{n}=2.776\cdot 2.17/\sqrt{5}=6.024/2.236=2.694$

Then, the lower and upper bounds of the CI are:

$LL=50.800-2.694=48.106\\\\UL=50.800+2.694=53.494$

The 95% confidence interval for the measurements is [48.106, 53.494].

4 0

4 years ago

Please help me pls and thank you!!

frosja888 [35]

Answer:

umm i don't see anything did i miss the joke lol

Step-by-step explanation:

3 0

3 years ago

0/1 Autumn's dog eats the same amount of dog food every day. On day 3, the bag of dog food has 12 pounds of dog food. On day 7,

erma4kov [3.2K]

Answer:

y - 0.5x

Step-by-step explanation:

Given that the dog eats the same amount of food each day and on day 3, the available quantity of food

= 12 pounds

On day 7 (4 days after), the available quantity of food is 10 pounds. The difference in the amount of food in 4 days

= 12 - 10

= 2 pounds

This means that each day, the amount of food eaten by the dog in one day

= 2/4

= 0.5 pound

Hence the amount of dog food in the bag y for any day after opening the dog food x will be given as

= y - 0.5x

6 0

3 years ago

A consumer advocacy group wants to determine whether there is a difference between the proportions of the two leading automobile

podryga [215]

Answer:

The answer is "0.3206".

Step-by-step explanation:

$H_0: p_1 = p_2\\\\H_a: p_1 \neq p_2\\\\\hat{p_1} = \frac{X_1}{N_1} = \frac{53}{400} = 0.1325\\\\\hat{p_1}= \frac{X_2}{N_2} = \frac{78}{500}= 0.156\\\\\hat{p} = \frac{(X_1 + X_2)}{(N_1 + N_2)} = \frac{(53+78)}{(400+500)} = 0.1456$

Testing statistic:

$z = \frac{(\hat{p_1}- \hat{p_2})}{\sqrt{(\hat{p} \times (1-\hat{p}) \times (\frac{1}{N_1} + \frac{1}{N_2}))}}$

$=\frac{(0.1325-0.156)}{\sqrt{(0.1456\times (1-0.1456)\times (\frac{1}{400} + \frac{1}{500}))}}\\\\ = -0.99$

Calculating the P-value Approach

$\text{P-value}= 0.3206$

7 0

3 years ago