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Roman55 [17]
3 years ago
11

How do you factor 7v2-47v+30

Mathematics
2 answers:
Semmy [17]3 years ago
5 0
7v² - 47v + 30

7v² - 5v - 42v + 30       split the second term into two terms.

v(7v - 5) - 6(7v - 5)      factor out common terms in the first two terms, then the last two terms.

(7v - 5)(v - 6)      << is your answer

hope this helps, God bless!
SVEN [57.7K]3 years ago
5 0
Or you can just use the X method
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Can anybody help on this (20 pts)
adell [148]
When x = -2, y = -5
When x = -1, y = -4
When x = 0, y = -3
When x = 1, y = -2
When x = 2, y = -1
Now plot these points on the graph: (-2,-5), (-1,-4), (0,-3), (1,-2), (2,-1)
8 0
2 years ago
The inside diameter of a randomly selected piston ring is a random variable with mean value 8 cm and standard deviation 0.03 cm.
S_A_V [24]

Answer:

a) P(7.99 ≤ X ≤ 8.01) = 0.8164

b) P(X ≥ 8.01) = 0.0475.

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 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.

Central limit theorem:

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

In this problem, we have that:

\mu = 8, \sigma = 0.03

(a) Calculate P(7.99 ≤ X ≤ 8.01) when n = 16.

n = 16, so s = \frac{0.03}{4} = 0.0075

This probability is the pvalue of Z when X = 8.01 subtracted by the pvalue of Z when X = 7.99. So

X = 8.01

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

Applying the Central Limit Theorem

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

Z = \frac{8.01 - 8}{0.0075}

Z = 1.33

Z = 1.33 has a pvalue of 0.9082

X = 7.99

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

Z = \frac{7.99 - 8}{0.0075}

Z = -1.33

Z = -1.33 has a pvalue of 0.0918

0.9082 - 0.0918 = 0.8164

P(7.99 ≤ X ≤ 8.01) = 0.8164

(b) How likely is it that the sample mean diameter exceeds 8.01 when n = 25? P(X ≥ 8.01) =

n = 25, so s = \frac{0.03}{5} = 0.006

This is 1 subtracted by the pvalue of Z when X = 8.01. So

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

Z = \frac{8.01 - 8}{0.006}

Z = 1.67

Z = 1.67 has a pvalue of 0.9525

1 - 0.9525 = 0.0475

P(X ≥ 8.01) = 0.0475.

4 0
3 years ago
(05.01)Which equation can be used to calculate the area of the shaded triangle in the figure below?
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aleksandrvk [35]
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y=-9+x^2
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y+9=x^2
square root both sides
\sqrt{y+9}=x

x=\sqrt{y+9}
5 0
3 years ago
Read 2 more answers
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