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3 years ago

I’m really confused on this problem.

Mathematics

1 answer:

marysya [2.9K]3 years ago

8 0

The area of the rectangle is 32 cm squared and the area of the triangle is 22 cm squared. When you add them together you get the area of the entire figure, which is 54 cm squared
Hope this helps!

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7 0

2 years ago

A gym teacher has $250 to spend on volleyball equipment. She buys 4 volleyball nets for $28 each. Volleyball cost $7 each. How m

Sonbull [250]

First find out how much the 4 volleyball nets are by using 4*28 which would equal to 112. Now subtract 112 from 250 and that leaves you with 138. Then you divide 138 by 7 which will give you about 19.7 which you need to take all decimals out since it isn’t an whole number. The gym teacher can only buy 19 volleyballs.

6 0

3 years ago

A professional baseball diamond is a square. The distance from base to base is 90 ft. To the nearest foot, how far does a catche

mamaluj [8]

BSE is square.

Let side be s

s=90ft

Diagonal be d

$\\ \tt\longmapsto s=\dfrac{d}{\sqrt{2}}$

$\\ \tt\longmapsto d=s\sqrt{2}$

$\\ \tt\longmapsto d=90\sqrt{2}$

$\\ \tt\longmapsto d=127ft$

3 0

2 years ago

Read 2 more answers

Rosetta wants to estimate the percentage of people who rent their home. She surveys 250 individuals and finds that 48 rent their

Andreas93 [3]

Answer:

$0.192 - 1.645\sqrt{\frac{0.192(1-0.192)}{250}}=0.151$

$0.192 + 1.645\sqrt{\frac{0.192(1-0.192)}{250}}=0.233$

Step-by-step explanation:

Information given

$X= 48$ number of people who rent their home

$n= 250$ represent the sample size

$\hat p =\frac{48}{250}= 0.192$ represent the proportion of people who rent their home

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=\pm 1.645$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.192 - 1.645\sqrt{\frac{0.192(1-0.192)}{250}}=0.151$

$0.192 + 1.645\sqrt{\frac{0.192(1-0.192)}{250}}=0.233$

3 0

3 years ago

A Ski resort tracks the proportion of seasonal employees who are rehired each season. Rehiring a seasonal employee is beneficial

DaniilM [7]

The test statistic z will be equal to -0.946 and it shows that there is no significant difference in the proportion of rehires between full time and part time.

Given sample sizes of 833 and 386 and result of samples 434 and 189.

$n_{1} =833, n_{2}=386$

Proportion of full time=434/833=0.52

Proportion of part time=189/386=0.49.

Difference in proportion =0.52-0.49

$H_{0}:$ TTF- i∈ rho=0

$H_{1}:$ TTF+i∈ rho≠0.

Mean of difference=0.03

Z=(X-μ)/σ

σ= $\sqrt{0.4517/447}$

=0.0317

σ=0.0317

z=(0-0.03)/0.0317

=-0.03/0.0317

=-0.317

p value will be =0.1736.

Because p value is greater than 0.01 so we will accept the null hypothesis which shows that there is no significant difference in the proportions.

Hence there is no significant difference in the proportion of rehires between full time and part time.

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6 0

2 years ago