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Arlecino [84]
3 years ago
6

(4x-25)=(2xt5) solve for x

Mathematics
1 answer:
MAXImum [283]3 years ago
3 0
Hello there.

<span>(4x-25)=(2xt5)


</span><span>25/<span><span>−<span>2<span>t5</span></span></span>+<span>4</span></span></span>
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Can you please help me. If you help me i will give you brainliest ( question 2-5)
vfiekz [6]

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5 0
3 years ago
-1/6 + 2/3 ( 9 + - 3/4) + - 1/2
Liono4ka [1.6K]
First simplify the section in the parenthesis. 
-1/6 + 2/3(8 1/4) + -1/2 
Then multiply 2/3 by 8 1/4.
-1/6 + 5 1/2 + -1/2 
Add -1/2 to 5 1/2. 
-1/6 + 5 
Add 5 to -1/6. 
4 5/6 is the fully simplified answer. 
Hope this helps!
7 0
3 years ago
The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 30,393 miles, with a standard
Pie

Answer:

52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

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 s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 30393, \sigma = 2876, n = 37, s = \frac{2876}{\sqrt{37}} = 472.81

What is the probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

This probability is the pvalue of Z when X = 30393 + 339 = 30732 subtracted by the pvalue of Z when X = 30393 - 339 = 30054. So

X = 30732

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

By the Central Limit Theorem

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

Z = \frac{30732 - 30393}{472.81}

Z = 0.72

Z = 0.72 has a pvalue of 0.7642.

X = 30054

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

Z = \frac{30054 - 30393}{472.81}

Z = -0.72

Z = -0.72 has a pvalue of 0.2358

0.7642 - 0.2358 = 0.5284

52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

4 0
3 years ago
Which equation is equivalent to the equation 5X+30=45?
nika2105 [10]

Answer:

its probably 45-30=5x

6 0
3 years ago
Mary ran 1/5 of 1 km each day. How many meters did she run from Monday to Wednesday?
lord [1]

Answer:

3/5 or 0.6 in decimal form

Step-by-step explanation:

All you have to do is multiply 1/5 x 3 for the three days Monday, Tuesday and Wednesday and that gets you 3/5 or as 1/5 of a km is 0.2 you have to multiply 0.2 x 3 and you get 0.6 which is equivalent to 3/5

5 0
3 years ago
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