Polygons ABCD and EFGH are similar. Find the length of CD.

A random sample of 38 wheel chair users were asked whether they preferred cushion type A or B, and 28 of them preferred type A w

-BARSIC- [3]

Answer:

The statement that cushion A is twice as popular as cushion B cannot be verified

Step-by-step explanation:

From the question we are told that:

Sample size n=38

Type a size A $X_a=28$

Type a size B $X_b=10$

Generally the probability of choosing cushion A P(a) is mathematically given by

$P(a)=\frac{28}{38}$

$P(a)=0.73$

Generally the equation for A to be twice as popular as B is mathematically given by

$P(b)+2P(b)=3P$

Therefore Hypothesis

$Null H_0: p \leq \frac{2P}{3P} \\Altenative H_A:p>\frac{2P}{3P}$

Generally the equation normal approx of p value is mathematically given by

$z=\frac{x-np_0-0.5}{\sqrt{np_0(1-p_0)} }$

$z=\frac{28-(38*2/3)_0-0.5}{\sqrt{38*2/3*1/3} }$

$z=0.75$

Therefore from distribution table

$Pvalue=1-\theta (0.75)$

$Pvalue=0.227$

Therefore there is no sufficient evidence to disagree with the Null hypothesis $H_0$

Therefore the statement that cushion A is twice as popular as cushion B cannot be verified

7 0

3 years ago

A random sample of n = 9 structural elements is tested for comprehensive strength. We know the true mean comprehensive strength

irinina [24]

Answer:

Probability that the sample mean comprehensive strength exceeds 4985 psi is 0.99999.

Step-by-step explanation:

We are given that a random sample of n = 9 structural elements is tested for comprehensive strength. We know the true mean comprehensive strength μ = 5500 psi and the standard deviation is σ = 100 psi.

Let $\bar X$ = sample mean comprehensive strength

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean comprehensive strength = 5500 psi

$\sigma$ = standard deviation = 100 psi

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the sample mean comprehensive strength exceeds 4985 psi is given by = P( $\bar X$ > 4985 psi)

P( $\bar X$ > 4985 psi) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{4985-5500}{\frac{100}{\sqrt{9} } }$ ) = P(Z > -15.45) = P(Z < 15.45)

= 0.99999

Since in the z table the highest critical value of x for which a probability area is given is x = 4.40 which is 0.99999, so we assume that our required probability will be equal to 0.99999.

4 0

3 years ago

Pleaseeeeeeee help me

lorasvet [3.4K]

Answer:

by my best guess, x = -5 (negative five)

Step-by-step explanation:

3 0

3 years ago

You are given a problem that involves multiplying a positive number and a negative number.

kenny6666 [7]

Answer:

-3 and -2

Step-by-step explanation:

A better explanation...

(positive)(positive)= positive

(positive)(negative)= negative

(negative)(negative)= positive

(negative)(positive)= negative

3 0

3 years ago

Read 2 more answers

Help me with my homework. i will heart correct

anzhelika [568]

Answer:

A is y-4 x-4

B is y-3 x-2

C is y-1 x-2

D is y-1 x-4

Step-by-step explanation:

look at where they are lined up from y to x

8 0

3 years ago