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allochka39001 [22]
3 years ago
9

PLEASE HELP!!! 70 POINTS FOR ONE QUESTION!!! RIGHT ANSWER ONLY!!!

Mathematics
2 answers:
34kurt3 years ago
5 0
Next step is
3x=16
Because you're doubling the answer
Harlamova29_29 [7]3 years ago
5 0

After

3x - 3 = 13

we'd add 3 to both sides and get

3x = 16

Answer: first choice


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Answer please........
Likurg_2 [28]

you have to show the graph

4 0
3 years ago
A manufacturer of tires wants to advertise a mileage interval that ex-cludes no more than 10% of the mileage on tires he sells.
Sergio039 [100]

Answer:

This mileage interval is from 30120 miles and higher.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

All he knows is that, for a large number of tires tested, the mean mileage was 25,000 miles, and the standard deviation was 4000 miles. This means that \mu = 25000, \sigma = 4000.

A manufacturer of tires wants to advertise a mileage interval that ex-cludes no more than 10% of the mileage on tires he sells. What interval wouldyou suggest?

The lower end of this interval is X when Z has a pvalue of 0.90.  That is Z = 1.28.

So

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

1.28 = \frac{X - 25000}{4000}

X - 25000 = 4000*1.28

X = 30120

This mileage interval is from 30120 miles and higher.

7 0
3 years ago
Suppose two $20 bills, three $10 bills, one $5 bill, and seven $1 bills are placed in a bag. If you were to pull a bill at rando
Korolek [52]

Answer:

Assuming that the $1 bill was pulled at random, then the expected value of the amount chosen is \frac{7}{13}.

Step-by-step explanation:

From the given question, the bag contains;

$1 bill = 7

$5 bill = 1

$10 bill = 3

$20 bill = 2

Total number of bills in the bag = 13

Pulling a bill at random, the bills would have an expected value as follows:

For $1 bill, the expected value = \frac{7}{13}

For $5 bill, expected value = \frac{1}{13}

For $10 bill, expected value = \frac{3}{13}

For $20 bill, the expected value = \frac{2}{13}

Assuming that the $1 bill was pulled at random, then the expected value of the amount chosen is \frac{7}{13}.

6 0
3 years ago
A strand of lights has 50 bulbs and eight of them are burned out what is the ratio of total numbers of the bulbs to the bulbs th
Strike441 [17]

Answer:

B.50:8

Step-by-step explanation:

6 0
3 years ago
4 ME FISHY BROTHERs :)
harkovskaia [24]

Answer:

542.6

Step-by-step explanation:

5 0
3 years ago
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