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Olegator [25]
3 years ago
6

hi guys the problem is i don't know how to use this so i will post this for fun lol i hope this help your day get better because

i have to nothing to question not at the moment right now am just vibing to my music :)
Mathematics
1 answer:
ValentinkaMS [17]3 years ago
4 0

Answer: thats fun i guess keep Vibin doe

Step-by-step explanation:

You might be interested in
My question is this can you help me​
jok3333 [9.3K]

Answer:

It's 1/4

Step-by-step explanation:

It's closer to 0 than the other fractions

4 0
2 years ago
Which of the following is true about a linear function?
UkoKoshka [18]
It has a constant rate of change
3 0
3 years ago
I need help with this one. It’s #6. :((
jonny [76]

Answer:

C

Step-by-step explanation:

f(x) = 2ˣ + 1

-f(x) = -(2ˣ) − 1

First, let's find the y-intercept.

-f(0) = -(2⁰) − 1 = -2

Only C can be correct.

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
This is the one!!!!!!!!!!!
sukhopar [10]
All you have to do is move the denominator on the other side, by multiplying.

a-10
------- = -9  
20 

multiply 20 x -9
= -180

then it becomes a-10=-180
                             +10  +10
a=-170
4 0
3 years ago
Read 2 more answers
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