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Liula [17]
3 years ago
5

CAN SOMEONE ANSWER ASAP PLEASE!!!!! WILL GIVE EXTRA POINTS

Mathematics
1 answer:
sdas [7]3 years ago
3 0
I'm really sorry if I'm wrong but

14*26=364 so he can go 364 miles for every full tank so his first 364 are covered since he starts with a full tank.

Step 2:
838/26*14 = <span>451
</span>
step 3:


He'd only have to stop once

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How to solve monomials
rewona [7]
You can't solvd monomials because they are just 1 term multiplication :
example: 2a , or 14
8 0
4 years ago
How do you find the quadratic equation for the roots 3 and 17?
Naily [24]
The roots are 3 and 17 so the distributed equation is

(x-3)(x-17)

distribute 

x²-3x-17x+51

x²-20x+51

I hope I've helped! 
6 0
3 years ago
For a sample of nequals37​, find the probability of a sample mean being less than 12 comma 751 or greater than 12 comma 754 when
Irina18 [472]

Answer:

50% probability of a sample mean being less than 12,751 or greater than 12,754

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples 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.

Central limit theorem:

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, which is also called standard error s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 12751, \sigma = 2.1, n = 37, s = \frac{2.1}{\sqrt{37}} = 0.3456

Find the probability of a sample mean being less than 12,751 or greater than 12,754

Less than 12,751

pvalue of Z when X = 12751.

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

By the Central Limit Theorem

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

Z = \frac{12751 - 12751}{0.3456}

Z = 0

Z = 0 has a pvalue of 0.5.

50% probability of the sample mean being less than 12,751.

Greater than 12,754

1 subtracted by the pvalue of Z when X = 12,754.

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

Z = \frac{12754 - 12751}{0.3456}

Z = 8.68

Z = 8.68 has a pvalue of 1

1 - 1 = 0

0% probability of the sample mean being greater than 12754

Less than 12,751 or greater than 12,754

50 + 0 =50

50% probability of a sample mean being less than 12,751 or greater than 12,754

6 0
3 years ago
Please help, need asap
Debora [2.8K]

Answer:

Choice 2

Step-by-step explanation:

8 0
3 years ago
You deposit $634 into an investment that earns 6% compound interest for 6 years. What is the balance at the end of the 6 years?
Irina18 [472]

Answer:

864.4

Step-by-step explanation: if you find what 6% of 634 is it is 38.4 so then you multiply 38.4 by 6 and you get 230.4 and then you add that to 634.

hope this helps :)

4 0
3 years ago
Read 2 more answers
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