In what place value will the first digit of the quotient be for 5.64 divided 15.

How many solutions does the equation −2a + 2a + 7 = 8 have? One Two None Infinitely many.

jonny [76]

The answer is none. Since you have to add your a's first, your left with 0 meaning that there is no more a's meaning that you have no solution. Hope this helps!

7 0

3 years ago

Read 2 more answers

Suppose the test scores for a college entrance exam are normally distributed with a mean of 450 and a s. d. of 100. a. What is t

svet-max [94.6K]

Answer:

a) 68.26% probability that a student scores between 350 and 550

b) A score of 638(or higher).

c) The 60th percentile of test scores is 475.3.

d) The middle 30% of the test scores is between 411.5 and 488.5.

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 450, \sigma = 100$

a. What is the probability that a student scores between 350 and 550?

This is the pvalue of Z when X = 550 subtracted by the pvalue of Z when X = 350. So

X = 550

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

$Z = \frac{550 - 450}{100}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 350

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

$Z = \frac{350 - 450}{100}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

68.26% probability that a student scores between 350 and 550

b. If the upper 3% scholarship, what score must a student receive to get a scholarship?

100 - 3 = 97th percentile, which is X when Z has a pvalue of 0.97. So it is X when Z = 1.88

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

$1.88 = \frac{X - 450}{100}$

$X - 450 = 1.88*100$

$X = 638$

A score of 638(or higher).

c. Find the 60th percentile of the test scores.

X when Z has a pvalue of 0.60. So it is X when Z = 0.253

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

$0.253 = \frac{X - 450}{100}$

$X - 450 = 0.253*100$

$X = 475.3$

The 60th percentile of test scores is 475.3.

d. Find the middle 30% of the test scores.

50 - (30/2) = 35th percentile

50 + (30/2) = 65th percentile.

35th percentile:

X when Z has a pvalue of 0.35. So X when Z = -0.385.

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

$-0.385 = \frac{X - 450}{100}$

$X - 450 = -0.385*100$

$X = 411.5$

65th percentile:

X when Z has a pvalue of 0.35. So X when Z = 0.385.

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

$0.385 = \frac{X - 450}{100}$

$X - 450 = 0.385*100$

$X = 488.5$

The middle 30% of the test scores is between 411.5 and 488.5.

7 0

3 years ago

Solve for X Need help ASAP

dezoksy [38]

X = nine. Seventy five / 9.75!

6 0

3 years ago

5. Which polygon is always regular?

Feliz [49]

Square because it's sides and angles are always congruent with eachother

7 0

2 years ago

A certain company sends 35% of its overnight mail parcels via express delivery service 1 and the rest by express delivery servic

sineoko [7]

Answer:

55.32% probability that a late package was delivered by express delivery service 2

Step-by-step explanation:

Bayes Theorem:

Two events, A and B.

$P(B|A) = \frac{P(B)*P(A|B)}{P(A)}$

In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.

In this question:

Event A: Late delivery.

Event B: Service 2 was used.

A certain company sends 35% of its overnight mail parcels via express delivery service 1 and the rest by express delivery service 2.

100 - 35 = 65%.

So $P(B) = 0.65$

Service 2 has a record of 2.0% of packages being delivered late.

This means that $P(A|B) = 0.02$

Probability of a late delivery.

35% from service 1. Of those, 3% are late.

65% from service 2. Of those, 2% are late.

So

$P(A) = 0.35*0.03 + 0.65*0.02 = 0.0235$

What is the probability that a late package was delivered by express delivery service 2

$P(B|A) = \frac{0.65*0.02}{0.0235} = 0.5532$

55.32% probability that a late package was delivered by express delivery service 2

5 0

2 years ago