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Licemer1 [7]
3 years ago
7

-2 +2-3(x-4) Complete the missing value in the solution to the equation

Mathematics
1 answer:
olga_2 [115]3 years ago
8 0
-2 + 2 = 0 0 - 3 = -3 -3 x x = -3x
-3 x -4 = -12
Answer : -3x - 12

I think
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Help me with this please<br> who ever is correct gets Brainliest
zimovet [89]

Answer:

ok it is there is no question lol

Step-by-step explanation:

4 0
3 years ago
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State the slope-intercept form of the equation for the graph with the points 3,0 and 0,-1
Dvinal [7]

Answer:

The answer is y=1/3x-1

Step-by-step explanation:

m=1/3

y-y=m(x-x1)

y-(-1)=1/3(x-0)

y+1=1/3x-0

-1 -1

y=1/3x-1

6 0
3 years ago
What is the solution to the equation 15=/x/ -5=
muminat

Answer:

This is absolute value equation so collect like terms and find X

15+5=|x|

20=|x|

since it's absolute value it's going to be 20 or -20

7 0
3 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
Can you guys help me please ASAP
Cerrena [4.2K]
B would be your best answer looking at the graphic chart
8 0
3 years ago
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