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nlexa [21]
3 years ago
7

What is the answer f(t)=40(1.18)^t

Mathematics
1 answer:
Scilla [17]3 years ago
5 0

Answer:

18% if it is the rate of change.

0 if solved

Step-by-step explanation:

Sorry, but you need more information

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Sweet tea purchased a new pair of jeans to wear to the battle of the bands on sale at 40% off.He saved 20 dollars.What was the r
MaRussiya [10]

Answer:

50

Step-by-step explanation:

50$ because 20$ is 40%, so times that by 2 which you get 20$ and 80%. Then you divide 20 by 2 then you get 10 which is 20% so you add 10 and you get 50$ which is 100%

3 0
3 years ago
Can someone help I’m so confused
erastova [34]

Answer: Use M A T H W A Y

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Someone help me with this please !!!!!:(
Zielflug [23.3K]

Answer:

8.1.

Step-by-step explanation:

a^2+b^2=c^2

8^2+2^2=c^2

64+4

68

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2 years ago
Which expression is represented by the phrase "the square of y decreased by the quotient of 28 and 7"?
padilas [110]

Answer:

y^2-(28/7)

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3 0
2 years ago
The tread life of a particular brand of tire is a random variable best described by a normal distribution with a mean of 60,000
Elis [28]

Answer:

84.13% probability a particular tire of this brand will last longer than 57,100 miles

Step-by-step explanation:

When the distribution is normal, we use 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 = 60000, \sigma = 2900

What is the probability a particular tire of this brand will last longer than 57,100 miles

This is 1 subtracted by the pvalue of Z when X = 57100. So

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

Z = \frac{57100 - 60000}{2900}

Z = -1

Z = -1 has a pvalue of 0.1587

1 - 0.1587 = 0.8413

84.13% probability a particular tire of this brand will last longer than 57,100 miles

7 0
3 years ago
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