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olya-2409 [2.1K]
3 years ago
8

17/15= 10/2x-2 . what is the value of x

Mathematics
1 answer:
dezoksy [38]3 years ago
7 0

Answer:

x=47/75

Step-by-step explanation:

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Find the perimeter of the image below
Naily [24]

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46

Step-by-step explanation:

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A certain ore is 31.8% nickel by mass. How many kilograms of this ore would you need to dig up to have 25.0 g of nickel?
natulia [17]
It will need 1.8 kilograms to have 25.0 g of nickel
6 0
3 years ago
10 to the power of 4 as a decimal
Nonamiya [84]

Answer:

You treat it as 10^4 (the small number is the number of zeros. Therefore the answer is 10000. Or of you meant 10^-4. Its the same concept, just reversed. So the answer in this case is 0.0004.

Cheers! :)

3 0
2 years ago
Need help with this question
Amiraneli [1.4K]

Answer:

1 and 3, 2 and 4

Step-by-step explanation:

Supplementary angles total to 180° or should form a straight line. 1 and 3, 2 and 4 are vertical angles which equal one another and are not supplementary.

6 0
3 years ago
A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is .5
lara [203]

Answer:

7.3% of the bearings produced will not be acceptable

Step-by-step explanation:

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 = 0.499, \sigma = 0.002

Target value of .500 in. A bearing is acceptable if its diameter is within .004 in. of this target value.

So bearing larger than 0.504 in or smaller than 0.496 in are not acceptable.

Larger than 0.504

1 subtracted by the pvalue of Z when X = 0.504.

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

Z = \frac{0.504 - 0.494}{0.002}

Z = 2.5

Z = 2.5 has a pvalue of 0.9938

1 - 0.9938= 0.0062

Smaller than 0.496

pvalue of Z when X = -1.5

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

Z = \frac{0.496 - 0.494}{0.002}

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

0.0668 + 0.0062 = 0.073

7.3% of the bearings produced will not be acceptable

4 0
3 years ago
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