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3 years ago

Please answer quickk

Mathematics

2 answers:

Usimov [2.4K]3 years ago

8 0

Answer:

t + 25 = 26

t = 26 - 25

t = 1

I hope it's helps you

aleksandrvk [35]3 years ago

7 0

T is 1. __ + 25 is 26. 1+25=26

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Talja [164]

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3 years ago

Red bricks and grey bricks were used to build a decorative wall. The number of red bricks divided by number of grey bricks was 5

azamat

Red bricks : Grey bricks
5 : 2

Total parts = 5 + 2 = 7

7 parts = 224
-----------------------
Find 1 part:
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1 part = 234 ÷ 7
1 part = 32

-----------------------
Find 5 parts:
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5 parts = 32 x 5
5 parts = 160

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Answer: 160 Red bricks
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4 years ago

PLZ HELP 20 POINTS

katrin2010 [14]

Can you tell us what the statement is and what we have to fill in please?

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3 years ago

Read 2 more answers

In a simple random sample of 14001400 young people, 9090% had earned a high school diploma. Complete parts a through d below.

ratelena [41]

Answer:

(a) The standard error is 0.0080.

(b) The margin of error is 1.6%.

(c) The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d) The percentage of young people who earn high school diplomas has increased.

Step-by-step explanation:

Let p = proportion of young people who had earned a high school diploma.

A sample of n = 1400 young people are selected.

The sample proportion of young people who had earned a high school diploma is:

$\hat p=0.90$

(a)

The standard error for the estimate of the percentage of all young people who earned a high school diploma is given by:

$SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}$

Compute the standard error value as follows:

$SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}$

$=\sqrt{\frac{0.90(1-0.90)}{1400}}\\$

$=0.008$

Thus, the standard error for the estimate of the percentage of all young people who earned a high school diploma is 0.0080.

(b)

The margin of error for (1 - α)% confidence interval for population proportion is:

$MOE=z_{\alpha/2}\times SE_{\hat p}$

Compute the critical value of z for 95% confidence level as follows:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the margin of error as follows:

$MOE=z_{\alpha/2}\times SE_{\hat p}$

$=1.96\times 0.0080\\=0.01568\\\approx1.6\%$

Thus, the margin of error is 1.6%.

(c)

Compute the 95% confidence interval for population proportion as follows:

$CI=\hat p\pm MOE\\=0.90\pm 0.016\\=(0.884, 0.916)\\\approx (88.4\%,\ 91.6\%)$

Thus, the 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d)

To test whether the percentage of young people who earn high school diplomas has increased, the hypothesis is defined as:

H₀: The percentage of young people who earn high school diplomas has not increased, i.e. p = 0.80.

Hₐ: The percentage of young people who earn high school diplomas has not increased, i.e. p > 0.80.

Decision rule:

If the 95% confidence interval for proportions consists the null value, i.e. 0.80, then the null hypothesis will not be rejected and vice-versa.

The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

The confidence interval does not consist the null value of p, i.e. 0.80.

Thus, the null hypothesis is rejected.

Hence, it can be concluded that the percentage of young people who earn high school diplomas has increased.

8 0

3 years ago

I need help!!! please help me!!!

natta225 [31]

Answer:

um ok the laiyflylfug;ou;

Step-by-step explanation:

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3 years ago