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wariber [46]
2 years ago
5

In 1990, the population of Math Valley was 15,000. If the population is increasing at an annual rate of 2.4%, what was the popul

ation in 1965?
Mathematics
1 answer:
Marizza181 [45]2 years ago
7 0

if this helped you say meow

i may be wrong but i think you should divide 2.4% by the 15000 wicth  is

0.00000016 or just 0.16 or maybe  muptication 2.4% x15000 whicth is 360 hope this helps :D

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Divide 33 into 2 groups so the ratio can be 4 to 7
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Your result is ok.
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Find the upper limit for the zeros of the function P(x)=4x^4+8x^3-7x^2-21x-9
pishuonlain [190]

Answer:

OPTION B - 2

Step-by-step explanation:

What is the upper limit for the zeros of the function P(x) = 4x^4 + 8x^3 - 7x^2 - 21x - 9. Ans: 2 is an upper limit. Use synthetic division. and the remainder are all positive, 2 is an upper limit.

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3 years ago
Can someone please help me?? :( I will give brainliest
Simora [160]

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Step-by-step explanation:

5 0
2 years ago
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A data set lists weights​ (lb) of plastic discarded by households. The highest weight is 5.31 ​lb, the mean of all of the weight
a_sh-v [17]

Answer:

a) The difference is of 3.222 lbs.

b) 1.64 standard deviations.

c) Z = 1.64

d) Not significant, as the z-score of 1.64 is between -2 and 2.

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.

The mean of all of the weights is x=2.088 ​lb, and the standard deviation of the weights is s=1.968 lb.

This means that \mu = 2.088, \sigma = 1.968

a. What is the difference between the weight of 5.31 lb and the mean of the​ weights?

This is X - \mu = 5.31 - 2.088 = 3.222

The difference is of 3.222 lbs.

b. How many standard deviations is that​ [the difference found in part​ (a)]?

This is the z-score. So

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

Z = \frac{5.31 - 2.088}{1.968}

Z = 1.64

1.64 standard deviations.

c. Convert the weight of 5.31 lb to a z score.

Z = 1.64, as found above.

d. If we consider weights that convert to z scores between −2 and 2 to be neither significantly low nor significantly​ high, is the weight of 5.31 lb​ significant?

Not significant, as the z-score of 1.64 is between -2 and 2.

5 0
3 years ago
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