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2 years ago

Can someone help meeeeee

Mathematics

1 answer:

11Alexandr11 [23.1K]2 years ago

7 0

Answer:

Area of shaded rectangle = 42 cm²

55%

Step-by-step explanation:

The length of the shaded area is equal to the length of the white rectangle.

Therefore, length of shaded rectangle = 7 cm

The width of the shaded rectangle is equal to the length of the white rectangle <u>minus</u> two widths of the white rectangle.

Therefore, width = 7 - (2 x 0.5) = 6 cm

Area of shaded rectangle = width x length

= 6 x 7

= 42 cm²

----------------------------------------------------------------------------------------------

Perimeter of a square = 4 × side length

If the perimeter of the square is 40 cm,
then the side length = 40 ÷ 4 = 10 cm

Area of a square = side length x side length

⇒ area of this square = 10 x 10 = 100 cm²

To determine the shaded area, calculate the areas of the 2 white triangles and subtract these from the area of the square.

Area of a triangle = 1/2 x base x height

⇒ area of left triangle = 1/2 x (10 - 7) x 10 = 15 cm²

⇒ area of right triangle = 1/2 x 10 x (10 - 4) = 30 cm²

Therefore, shaded area = 100 - 15 - 30 = 55 cm²

To calculate the percentage, divide the shaded area by the area of the square and multiply by 100:

(55 ÷ 100) × 100% = 55%

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Company A makes a large shipment to Company B. Company B can reject the shipment if they can conclude that the proportion of def

boyakko [2]

Answer:

$z=\frac{0.1475-0.1}{\sqrt{\frac{0.1(1-0.1)}{400}}}=3.17$

The p value for this case would be given by:

$p_v =P(z>3.17)=0.00076$

For this case the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 0.1 and then Company B can reject the shipment

Step-by-step explanation:

Information provided

n=400 represent the random sample taken

X=59 represent number of defectives from the company B

$\hat p=\frac{59}{400}=0.1475$ estimated proportion of defectives from the company B

$p_o=0.1$ is the value to verify

$\alpha=0.05$ represent the significance level

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to verify if the true proportion of defectives is higher than 0.1 then the system of hypothesis are.:

Null hypothesis: $p \leq 0.1$

Alternative hypothesis: $p > 0.1$

The statistic would be given by:

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

Replacing the info given we got:

$z=\frac{0.1475-0.1}{\sqrt{\frac{0.1(1-0.1)}{400}}}=3.17$

The p value for this case would be given by:

$p_v =P(z>3.17)=0.00076$

8 0

3 years ago

Help Plz !!!

coldgirl [10]

The answer is A i think

7 0

2 years ago

A study was conducted to determine whether magnets were effective in treating pain. The values represent measurements of pain us

Zanzabum

Answer:

Step-by-step explanation:

Hello!

The objective of the study is to determine whether the magnets are effective in treating pain. For these two independent groups of individuals with equal affections were randomly sampled, one was treated with magnets and the other group of individuals, call it control group, was treated with a placebo treatment.

The information for both samples is:

X₁: pain measurement using a visual analog scale after the individual received magnet treatments.

X₁~N(μ₁;σ₁²)

n₁= 20

X[bar]₁= 0.46

S₁= 0.93

X₂: pain measurement using a visual analog scale of an individual of the control groups.

X₂~N(μ₂;σ₂²)

n₂= 20

X[bar]₂= 0.41

S₂= 1.26

The claim is that the pain reductions of the control group have more variation than the pain reductions of the target group. If it's so then we could suspect that the population variance of the control group, σ₂², will be greater than the population variance of the magnet group, σ₁².

To test the relationship between these two population variances you have to conduct a variance ratio test using the Snedecors F statistic.

The hypotheses are:

H₀: σ₂² ≤ σ₁²

H₁: σ₂² > σ₁²

α: 0.05

$F= (\frac{S_2^2}{S_1^2}) * (\frac{Sigma_2^2}{Sigma_1^2} )~~ F_{(n_2-1);(n_1-1)}$

This hypothesis test is one-tailed to the right, there is only one critical value:

$F_{(n_2-1);(n_1-1); 1 - \alpha } = F_{19;19; 0.95}= 2.17$

The decision rule for the test is:

If $F_{H_0}$ ≥ 2.17, then the decision is to reject the null hypothesis.

If $F_{H_0}$ < 2.17, then the decision is to not reject the null hypothesis.

$F_{H_0}= \frac{S_2^2}{S_1^2}*\frac{Sigma_2^2}{Sigma_1^2} = \frac{(1.26)^2}{(0.93)^2} * 1$

$F_{H_0} = 1.835 = 1.84$

Since the calculated value of F is less than the critical value, the decision is to reject the null hypothesis. So using a 5% significance level the decision is to not reject the null hypothesis, you can conclude that the population variance of the pain reduction of the control group is less or equal than the population variance of the pain reduction on individuals treated with magnets.

I hope it helps!

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