Please help!!!

To original: 9.6 new: 5.9 explain weather it decrease or increased

mixas84 [53]

Decreased because it was 9.6 and now it is 5.9

6 0

3 years ago

Read 2 more answers

Add my s.n.a.p clown6941

mina [271]

Answer:

lol

Step-by-step explanation:

3 0

3 years ago

Mrs Rahlee bought 300 yards of ribbon to make flowers. She used 1.22 yards to make one large flower. She made 200 such large flo

Mamont248 [21]

So 1.22 per large flower
200 large flowers
1.22 times 200=244 yards used
subtract
300-244=56

56=100 smal flowers
divide both sides by 100
0.56=1 small flower
the answer is 0.56 yards=1 flower

7 0

3 years ago

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What is 6/5 x 25/24 = ?

Margaret [11]

The answer is 5/4 or 1 and 1/4 (both answers are correct just different forms), hope this helps !

5 0

3 years ago

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In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s

Oksi-84 [34.3K]

Answer:

a) $\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

b) From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

c) $P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

$P(Z\geq2.070)=1-P(Z$

Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

$\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4$

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

$\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

Part b

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

Part c

For this case we want this probability:

$P(\bar X \geq 285)$

And we can use the z score defined as:

$z=\frac{\bar x -\mu}{\sigma_{\bar x}}$

And using this we got:

$P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

And using a calculator, excel or the normal standard table we have that:

$P(Z\geq2.070)=1-P(Z$

8 0

3 years ago