Use synthetic division and the Remainder Theorem to find P(a).

At 2 pm on Tuesday, the temperature was -12 degrees. By 6 pm the temperature had dropped to -20 degrees. What was the average ch

fiasKO [112]

Answer:

a

Step-by-step explanation:

6 0

3 years ago

Help plz!! 100 points

Fynjy0 [20]

Answer:

t = 0, 1, 4, 8

h = 0, 12, 0 , 18

Step-by-step explanation:

t = 0 + 1 + 3 + 4

/ x1 - 5 (2) x + 4x + 9

h = 0, 12, 0, 18

After 18 seconds the ball attains...

x/3 = 1/4 - 4x 2(7) = 9/2 - 2 / 4 (0)

6 0

3 years ago

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The mean work week for engineers in start-up companies is claimed to be about 64 hours with a standard deviation of 8 hours. Kar

blondinia [14]

Answer:

1.) mean

2.) H0 : μ = 64

3.) 0.0028

4) Yes

Step-by-step explanation:

Null hypothesis ; H0 : μ = 64

Alternative hypothesis ; H1 : μ < 64

From the data Given :

70; 45; 55; 60; 65; 55; 55; 60; 50; 55

Using calculator :

Xbar = 57

Sample size, n = 10

Standard deviation, s = 7.14

Test statistic :

(xbar - μ) ÷ s/sqrt(n)

(57 - 64) ÷ 8 / sqrt(10)

Test statistic = - 2.77

Pvalue = (Z < - 2.77) = 0.0028 ( Z probability calculator)

α = 10% = 0.1

Reject H0 ; if P < α

Here,

P < α ; Hence, we reject the null

6 0

3 years ago

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean

Mamont248 [21]

Answer:

a) The percentage of snails that take more than 60 hours to finish is 4.75%.

b) The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c) The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d) 0% probability that a randomly-chosen snail will take more than 76 hours to finish

e) To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f) The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

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:

$\mu = 50, \sigma = 6$

a. The percentage of snails that take more than 60 hours to finish is

This is 1 subtracted by the pvalue of Z when X = 60.

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

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

1 - 0.9525 = 0.0475

The percentage of snails that take more than 60 hours to finish is 4.75%.

b. The relative frequency of snails that take less than 60 hours to finish is

This is the pvalue of Z when X = 60.

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

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c. The proportion of snails that take between 60 and 67 hours to finish is

This is the pvalue of Z when X = 67 subtracted by the pvalue of Z when X = 60.

X = 67

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

$Z = \frac{67 - 50}{6}$

$Z = 2.83$

$Z = 2.83$ has a pvalue 0.9977

X = 60

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

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

0.9977 - 0.9525 = 0.0452

The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d. The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 1 subtracted by the pvalue of Z when X = 76.

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

$Z = \frac{76 - 50}{6}$

$Z = 4.33$

$Z = 4.33$ has a pvalue of 1

1 - 1 = 0

0% probability that a randomly-chosen snail will take more than 76 hours to finish

e. To be among the 10% fastest snails, a snail must finish in at most hours.

At most the 10th percentile, which is the value of X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

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

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f. The most typical 80% of snails take between and hours to finish.

From the 50 - 80/2 = 10th percentile to the 50 + 80/2 = 90th percentile.

10th percentile

value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

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

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

90th percentile.

value of X when Z has a pvalue of 0.9. So X when Z = 1.28

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

$1.28 = \frac{X - 50}{6}$

$X - 50 = 1.28*6$

$X = 57.68$

The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

5 0

4 years ago

A father and his daughter together weigh seven times the daughter's weight x. The father weighs 180 lbs. Which equation can be u

bezimeni [28]

Father and daguther together
father=180
father and daughter is 7 times daught
father+x=7x
180+x=7x
D is answer
and daughter=30lb

3 0

3 years ago

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