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2 years ago

HURRY NEED HELP ASAP

Mathematics

2 answers:

poizon [28]2 years ago

5 0

I=p*r*t
The answer is 1131.20

vovikov84 [41]2 years ago

3 0

Answer:

1,131.20

Step-by-step explanation:

I=prt

1,010*0.14*8

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Three less than the quotient of ten and a number, increased by six” when n = 2?

Musya8 [376]

(10/n +6) -3
10/2+6-3
5+6-3
11-3
8
Hope this helped!

3 0

3 years ago

Read 2 more answers

What is the square root of 9/16, 9 to the power of two

lbvjy [14]

Answer:

$\frac{3}{4}$ , 9

Step-by-step explanation:

$\sqrt{9/16}$ = $\sqrt{9}$ ÷ $\sqrt{16}$ = $\frac{3}{4}$

$\sqrt{9^2}$ = $\sqrt{81}$ = 9

3 0

3 years ago

Can someone plss help mee ??

Dvinal [7]

First - 8
second - 4
third - 12
fourth - 9

3 0

3 years ago

tim spends 25% of his time at the gym dooing leg exercises , 1/5 on shoulder exercises , 0.35 on biceps exercises , and warming

svetoff [14.1K]

Answer:

20%

Step-by-step explanation:

1/5 of 1 hour = 20 minutes

so 1/5 = 20%

25% + 20% = 45%

0.35 = 35%

45% + 35% = 80%

100% - 80% = 20%

Answer = 20%

Hope this helps!

3 0

3 years ago

A study was recently conducted at a major university to estimate the difference in the proportion of business school graduates w

sveta [45]

Answer:

$(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412$

$(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318$

And the 95% confidence interval would be given (-0.1412;-0.0318).

We are confident at 95% that the difference between the two proportions is between $-0.1412 \leq p_A -p_B \leq -0.0318$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_A$ represent the real population proportion for business

$\hat p_A =\frac{75}{400}=0.1875$ represent the estimated proportion for Business

$n_A=400$ is the sample size required for Business

$p_B$ represent the real population proportion for non Business

$\hat p_B =\frac{137}{500}=0.274$ represent the estimated proportion for non Business

$n_B=500$ is the sample size required for non Business

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412$

$(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318$

And the 95% confidence interval would be given (-0.1412;-0.0318).

We are confident at 95% that the difference between the two proportions is between $-0.1412 \leq p_A -p_B \leq -0.0318$

7 0

3 years ago