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sattari [20]
3 years ago
12

Need help! Please answer quickly! :)

Mathematics
1 answer:
guapka [62]3 years ago
4 0

Answer: a

Step-by-step explanation:

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Can someone please help i posted this question already and it got ingored
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The answer is C, since the number in the parenthesis represent what is being multiplied every year
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Graph g(x) = |x-4| +3
statuscvo [17]
The answer is (x77)(272)
7 0
3 years ago
Mr. Gonzazales has only 42.50 to spend he wants to buy 29 t shirts including tax and some bracelets that cost 4.50 each includin
erica [24]

Answer:

He can buy <u>3 bracelets</u>.

Step-by-step explanation:

Given:

Mr. Gonzales has only 42.50 to spend he wants to buy 29 t shirts including tax and some bracelets that cost 4.50 each including tax.

Now, to find the number of bracelets he can buy.

Let the number of bracelets he can buy be x.

Price of each bracelets = 4.50.

Total amount to spend = 42.50.

Number of t-shirts = 29.

Now, to get the number of bracelets we put an equation:

29+4.50\times x=42.50\\\\29+4.50x=42.50

<em>Subtracting both sides by 29 we get:</em>

<em />4.50x=42.50-29\\\\4.50x=13.5<em />

<em>Dividing both sides by 4.50 we get:</em>

x=3.

<u>The number of bracelets = 3.</u>

Therefore, he can buy 3 bracelets.

6 0
3 years ago
What are the X and Y intercepts of the line in the graph?
vovikov84 [41]

Answer:

Y-int: (0,3)

X-int: (4,0)

Equation: y = -3/4x + 3

Step-by-step explanation:

3 0
2 years ago
. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an
Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

Normal probability distribution

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:

\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12

What is the probability that the average score of a random sample of 20 students exceeds 59.5?

This is 1 subtracted by the pvalue of Z when X = 59.5. So

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

Z = \frac{59.5 - 56}{1.12}

Z = 3.1

Z = 3.1 has a pvalue of 0.9990.

So there is a 1-0.9990 = 0.001 = 0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

3 0
3 years ago
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