Find the zeros of the quadratic function

Three scholars each brought bags of skittles at the store that had exactly the same amount in each. Genesis ate 3/6 of a bag of

vlada-n [284]

Answer:

7/6

Step-by-step explanation:

add all numerators together

8 0

3 years ago

In a lab experiment, 2200 bacteria are placed in a petri dish. The conditions are such that the number of bacteria is able to do

ratelena [41]

The answer would be 5866 because after it multiplied twice it would be at 8800 at that rate it would be gaining 1466.66 an hour so add that to 4400 and u get 5866

6 0

3 years ago

Solve. -1/3b = 9<br> A.-3<br> B.3<br> C.-27<br> D.27

STALIN [3.7K]

Ok so lets plug in a first -1/3(-3) would equal 9 a is the accurate answer

3 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

Midpoint of (-2,-1) (-6,12)

Leni [432]

Answer:

(-4, 5.5)

Step-by-step explanation:

1. put your Xs and Ys over 2

2. add them together

3. Divide

4. Simplify

5. Turn fraction into decimal

$m(x.y) \\ m = \frac{x1 + x2}{2} . \: \frac{y1 + y2}{2} \\ m = \: \frac{ - 2 + - 6}{2} . \frac{ - 1 + 12}{2} \\ m = ( - \frac{8}{2} . \frac{11}{2} ) \\ m = ( - 4 \: . \: 5 \frac{1}{2} ) \\ \\ \\ mid = ( - 4 \: \: . \: 5.5)$

6 0

2 years ago