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3 years ago

I need help on pages 18 3-7 I’ll give you brainliest

Mathematics

1 answer:

Basile [38]3 years ago

6 0

Take a pic of it or look it up on safari or download slander it has ALOT of math book answers like this one

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Lesledi has a box that is 36cm wide, 24cm high, and 20cm deep. She wants to fill the box with packages that are 12cm wide, 6cm h

Paul [167]

Answer:

The answer is 3

Step-by-step explanation:

36÷12=3, which is the lowest number on either side of the boxes

5 0

3 years ago

Crash testing is a highly expensive procedure to evaluate the ability of an automobile to withstand a serious accident. A simple

Veseljchak [2.6K]

Answer:

The 95% confidence interval is (-0.2451, 06912)

Step-by-step explanation:

From the question, we have;

The number of small cars in the sample of small cars, n₁ = 12

The number of small cars that were totaled, x = 8

The number of large cars in the sample of small cars, n₂ = 15

The number of large cars that were totaled, y = 5

Therefore, the proportion of small cars that were totaled, pX = x/n₁

∴ pX = 8/12 = 2/3

The proportion of large cars that were totaled, pY = y/n₁

∴ pY = 5/15 = 1/3

The 95% confidence interval for the difference pX - pY is given as follows;

$pX-pY\pm z^{*}\sqrt{\dfrac{pX\left (1-pX \right )}{n_{1}}+\dfrac{pY\left (1-pY \right )}{n_{2}}}$

$\dfrac{2}{3} -\dfrac{1}{3} \pm 1.96 \times \sqrt{\dfrac{\dfrac{2}{3} \times \left (1-\dfrac{2}{3} \right )}{12}+\dfrac{\dfrac{1}{3} \times \left (1-\dfrac{1}{3} \right )}{15}}$

Therefore, we have;

$\therefore 95\% \ CI = \dfrac{1}{3} \pm 0.3578454$

The 95% confidence interval, CI = (-0.2451, 06912)

6 0

2 years ago

Solve the given problem using Venn Diagram. Show your solution

Troyanec [42]

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7 0

2 years ago

Which data would most likely show a negative relationship when graphed on a scatterplot?

IRINA_888 [86]

The correct answer is C . Outside temperature

5 0

3 years ago

If we wanted to create a new 90% confidence interval from a different sample for the proportion of those with a two on one date

slava [35]

Answer: 271

Step-by-step explanation:

The formula we use to find the sample size is given by :-

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$

, where $z_{\alpha/2}$ is the two-tailed z-value for significance level of $(\alpha)$

p = prior estimation of the proportion

E = Margin of error.

If prior estimation of the proportion is unknown, then we take p= 0.5 , the formula becomes

$n=0.5(1-0.5)(\dfrac{z_{\alpha/2}}{E})^2$

$n=0.25(\dfrac{z_{\alpha/2}}{E})^2$

Given : Margin of error : E= 0.05

Confidence level = 90%

Significance level $\alpha=1-0.90=0.10$

Using z-value table , Two-tailed z-value for significance level of $0.10$

$z_{\alpha/2}=1.645$

Then, the required sample size would be :

$n=0.25(\dfrac{1.645}{0.05})^2$

Simplify,

$n=270.6025\approx271$

Hence, the required minimum sample size =271

3 0

3 years ago