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3 years ago

13

Entering 38.00 into the Price of Sneakers field Entering 6.00 into the Price field Entering 3.00 into the Price of Leather field

True or False: You will no

1 answer:

eimsori [14]3 years ago

8 0

Answer:

This question seems incorrect.

Kindly take a look again and re-state it properly to enable me give the most accurate answer.

Thank you

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Ariel left her home at 9 a.m., driving at 45 mph. At 11 a.m. her brother left the home and started after her at 60 mph. At what

Vlada [557]

Answer:

5:00 pm

Step-by-step explanation:

Let

x -----> the umber of hours

y ----> the total distance in miles

we know that

The distance is equal to the speed multiplied by the time

Find out the distance traveled by Ariel at 11:00 am

11:00 am-09:00 am=2 hours

45(2)=90 miles

so

<em>Ariel's linear equation is</em>

$y=45x+90$ ----> equation A

<em>Ariel's brother linear equation is</em>

$y=60x$ ----> equation B

Equate equation A and equation B

$60x=45x+90$

Solve for x

$60x-45x=90$

$15x=90$

$x=6\ hours$

Adds 11:00 am + 6 hours=17:00 =5:00 pm

6 0

4 years ago

Serjik [45]

Answer

B

explanation

3480ones = 3480x1=3480

8 0

2 years ago

Evaluate. 9.75 + 4.625 + (-9.75) + 2.6=?

marusya05 [52]

The answer would be 7.225.

4 0

3 years ago

Read 2 more answers

Solve the equation 24-45t=16

I am Lyosha [343]

Subtract 24 from both sides (-45t=-8) then divide both sides by -45 (t=0.177)

7 0

4 years ago

Suppose the weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams. The weights

user100 [1]

Answer:

a) 0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b) The weight that 80% of the apples exceed is of 78.28g.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.

This means that $\mu = 85, \sigma = 8$

a. Find the probability a randomly chosen apple exceeds 100 g in weight.

This is 1 subtracted by the p-value of Z when X = 100. So

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

$Z = \frac{100 - 85}{8}$

$Z = 1.875$

$Z = 1.875$ has a p-value of 0.9697

1 - 0.9696 = 0.0304

0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b. What weight do 80% of the apples exceed?

This is the 100 - 80 = 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.

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

$-0.84 = \frac{X- 85}{8}$

$X - 85 = -0.84*8$

$X = 78.28$

The weight that 80% of the apples exceed is of 78.28g.

5 0

3 years ago