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3 years ago

5

Could ya’ll play my game? Only two people are playing, and it’s a roleplay game, which is pretty bad lol. https://scratch.mit.ed

u/projects/595391965/ I’d really appreciate it if you did! :)

2 answers:

IRISSAK [1]3 years ago

7 0

Thx for the points! I am in school rn, im not getting on scratch.

Naily [24]3 years ago

5 0

Answer:

Ok

Step-by-step explanation:

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hello hello the area of a triangle is 32 square in the height of the triangle is 8 in what is the length of the base of a triang

solniwko [45]

As we know
area of triangle = 1/2 *height*base
32=1/2*8*base
32=4*base
8 =base

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3 years ago

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If (54)m = 512, what is the value of m?

Snezhnost [94]

The correct answer is 9.481.
Hope this helps.
Brainliest would be appreciated.

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Is 1.5 greater than 3

Andre45 [30]

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3 years ago

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the mgf of a random variable x is e^3(e^t-1). Find P[mean - standard deviation squared < X < 1/2( mean + standard deviatio

postnew [5]

The given MGF is that for a random variable following a Poisson distribution with parameter $\lambda=3$ .

This means $\mathbb E(X)=\mathbb V(X)=\lambda$ , and $X$ has PMF

$f_X(x)=\begin{cases}\dfrac{3^xe^{-3}}{x!}&\text{for }x\ge0\\0&\text{otherwise}\end{cases}$

So, the desired probability is

$\mathbb P\left(\lambda-\lambda^2$

This is equivalent to

$\displaystyle\sum_{x=0}^2\mathbb P(X=x)=\sum_{x=0}^2\frac{3^x}{x!e^3}=\frac{17}{2e^3}\approx0.4232$

8 0

3 years ago

A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. a) E

polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) $\alpha =1-0.98=0.02$

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

$p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }$

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and $z_{\alpha/2}$ is the z-value that let a probability of $\alpha/2$ on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, $\alpha$ by 0.02 and $z_{\alpha/2}$ by 2.33

Where p' and $\alpha$ are calculated as:

$p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02$

So, replacing the values we get:

$0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438$

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence $1-\alpha$ is equal to 98%

It means the the level of significance $\alpha$ is:

$\alpha =1-0.98=0.02$

4 0

3 years ago