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Soloha48 [4]
3 years ago
14

Using the graph below, if f(x) = 4, find x.

Mathematics
1 answer:
Rina8888 [55]3 years ago
4 0
3 I think is the answer
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Find the rate and unit rate:
Alenkinab [10]



A: 24 miles per 1 gallon
5 0
3 years ago
Read 2 more answers
What percentage of the questions did he get right
Wittaler [7]

Answer: 75% 45/60 is .75 witch is 75%

Step-by-step explanation:

8 0
3 years ago
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A family is planning a three-week vacation for which they will drive across the country. They have a van that gets 12 miles per
svetoff [14.1K]

Answer:

The family will pay 34722 cents more if they take the van

Step-by-step explanation:

Given

Van = 12 miles per gallon

Sedan = 36 miles per gallon

Distance = 2500 miles

Gas = $2.50 per gallon

First, we need to determine the number of gallons that'll be used by both vehicles

This  is done by dividing total distance by number of miles per gallon

Van = \frac{2500}{12} \ gallon

Sedan = \frac{2500}{36}\ gallon

Next, is to multiply this by the cost of gas per gallon;

This gives the total spendable amount on both vehicles

Van = \frac{2500}{12} * \$2.50

Van = \frac{\$6250}{12}

Van = \$520.833

Sedan = \frac{2500}{36}* \$2.50

Sedan = \frac{\$6250}{36}

Sedan = \$173.611

Next is to get the difference between these amounts

Difference = \$520.833 - \$173.611

Difference = \$347.222

Multiply by 100 to convert to cents

Difference = 347.222 * 100 cents

Difference = 34722.2 \ cents

Difference = 34722\ cents (Approximated)

Hence;

<em>The family will pay 34722 cents more if they take the van</em>

8 0
3 years ago
Timothy put $300 in a bank that pays 7% interest rate, if he leaves his interest from the first year in the bank, how much money
andriy [413]

Answer: he would have $343.47 after 2 years.

Step-by-step explanation:

if he leaves his interest from the first year in the bank, we would assume that his interest was compounded. We would apply the formula for determining compound interest which is expressed as

A = P(1+r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = $300

r = 7% = 7/100 = 0.07

n = 1 because it was compounded once in a year.

t = 2 years

Therefore,.

A = 300(1 + 0.07/1)^1 × 2

A = 300(1.07)^2

A = $343.47

7 0
3 years ago
The distribution of lifetimes of a particular brand of car tires has a mean of 51,200 miles and a standard deviation of 8,200 mi
Orlov [11]

Answer:

a) 0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

b) 0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

c) 0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

d) 0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

Step-by-step explanation:

Problems of normally distributed distributions 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 question, we have that:

\mu = 51200, \sigma = 8200

Probabilities:

A) Between 55,000 and 65,000 miles

This is the pvalue of Z when X = 65000 subtracted by the pvalue of Z when X = 55000. So

X = 65000

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

Z = \frac{65000 - 51200}{8200}

Z = 1.68

Z = 1.68 has a pvalue of 0.954

X = 55000

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

Z = \frac{55000 - 51200}{8200}

Z = 0.46

Z = 0.46 has a pvalue of 0.677

0.954 - 0.677 = 0.277

0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

B) Less than 48,000 miles

This is the pvalue of Z when X = 48000. So

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

Z = \frac{48000 - 51200}{8200}

Z = -0.39

Z = -0.39 has a pvalue of 0.348

0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

C) At least 41,000 miles

This is 1 subtracted by the pvalue of Z when X = 41,000. So

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

Z = \frac{41000 - 51200}{8200}

Z = -1.24

Z = -1.24 has a pvalue of 0.108

1 - 0.108 = 0.892

0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

D) A lifetime that is within 10,000 miles of the mean

This is the pvalue of Z when X = 51200 + 10000 = 61200 subtracted by the pvalue of Z when X = 51200 - 10000 = 412000. So

X = 61200

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

Z = \frac{61200 - 51200}{8200}

Z = 1.22

Z = 1.22 has a pvalue of 0.889

X = 41200

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

Z = \frac{41200 - 51200}{8200}

Z = -1.22

Z = -1.22 has a pvalue of 0.111

0.889 - 0.111 = 0.778

0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

4 0
3 years ago
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