WORTH 30 POINTS PLZZZZZZZZZZZ HELP

What is 3,542 divide by 154

liubo4ka [24]

Answer:

23

Good Luck!!!

8 0

3 years ago

Suppose that the national average for the math portion of the College Board's SAT is 515. The College Board periodically rescale

nasty-shy [4]

Answer:

a) 16% of students have an SAT math score greater than 615.

b) 2.5% of students have an SAT math score greater than 715.

c) 34% of students have an SAT math score between 415 and 515.

d) $Z = 1.05$

e) $Z = -1.10$

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the empirical rule.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Empirical rule

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

$\mu = 515, \sigma = 100$

(a) What percentage of students have an SAT math score greater than 615?

615 is one standard deviation above the mean.

68% of the measures are within 1 standard deviation of the mean. The other 32% are more than 1 standard deviation from the mean. The normal probability distribution is symmetric. So of those 32%, 16% are more than 1 standard deviation above the mean and 16% more then 1 standard deviation below the mean.

So, 16% of students have an SAT math score greater than 615.

(b) What percentage of students have an SAT math score greater than 715?

715 is two standard deviations above the mean.

95% of the measures are within 2 standard deviations of the mean. The other 5% are more than 2 standard deviations from the mean. The normal probability distribution is symmetric. So of those 5%, 2.5% are more than 2 standard deviations above the mean and 2.5% more then 2 standard deviations below the mean.

So, 2.5% of students have an SAT math score greater than 715.

(c) What percentage of students have an SAT math score between 415 and 515?

415 is one standard deviation below the mean.

515 is the mean

68% of the measures are within 1 standard deviation of the mean. The normal probability distribution is symmetric, which means that of these 68%, 34% are within 1 standard deviation below the mean and the mean, and 34% are within the mean and 1 standard deviation above the mean.

So, 34% of students have an SAT math score between 415 and 515.

(d) What is the z-score for student with an SAT math score of 620?

We have that:

$\mu = 515, \sigma = 100$

This is Z when X = 620. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{620 - 515}{100}$

$Z = 1.05$

(e) What is the z-score for a student with an SAT math score of 405?

We have that:

$\mu = 515, \sigma = 100$

This is Z when X = 405. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{405 - 515}{100}$

$Z = -1.10$

3 0

3 years ago

What would a 64 ring size be in the US THANKSSSS

Mazyrski [523]

Answer:

10.75 would be a 64 ring size in the US

5 0

2 years ago

I need Help please!!

PSYCHO15rus [73]

I think it’s A because the s is between P&Q and S&C if it’s wrong I’m sorry!

3 0

2 years ago

The pesticide diazinon is in common use to treat infestations of the German cockroach, Blattella germanica. A study investigated

cluponka [151]

Answer:

We conclude that there is no difference in the proportion of cockroaches that died on each surface.

Step-by-step explanation:

We are given that a study investigated the persistence of this pesticide on various types of surfaces.

After 14 days, they randomly assigned 72 cockroaches to two groups of 36, placed one group on each surface, and recorded the number that died within 48 hours. On the glass, 18 cockroaches died, while on plasterboard, 25 died.

Let $p_1$ = proportion of cockroaches that died on glass surface.

$p_2$ = proportion of cockroaches that died on plasterboard surface.

So, Null Hypothesis, $H_0$ : $p_1-p_2$ = 0 {means that there is no difference in the proportion of cockroaches that died on each surface}

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq$ 0 {means that there is a significant difference in the proportion of cockroaches that died on each surface}

The test statistics that would be used here Two-sample z proportion statistics;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of cockroaches that died on glass surface = $\frac{18}{36}$ = 0.50

$\hat p_2$ = sample proportion of cockroaches that died on plasterboard surface = $\frac{25}{36}$ = 0.694

$n_1$ = sample of cockroaches on glass surface = 36

$n_2$ = sample of cockroaches on plasterboard surface = 36

So, test statistics = $\frac{(0.50-0.694)-(0)}{\sqrt{\frac{0.50(1-0.50)}{36}+\frac{0.694(1-0.694)}{36} } }$

= -1.712

The value of z test statistics is -1.712.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no difference in the proportion of cockroaches that died on each surface.

8 0

3 years ago