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KiRa [710]
4 years ago
14

A number g is decreased by 23 and then

Mathematics
1 answer:
iragen [17]4 years ago
6 0
A. -13

The equation would be

(G-23)1/2=2g+8

Multiply by 2 on both sides to get rid of the fraction

G-23=4g+16

subtract g from both sides

-23=3g+16

Subtract 16 from both sides

-39=3g

Divide 3 from both sides

g=-13
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See the picture attached to better understand the problem

we know that

the <span>smallest face of the prism is the face ABCD
</span>so
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6 0
3 years ago
Consider a value to be significantly low if its z score less than or equal to minus−2 or consider a value to be significantly hi
katrin2010 [14]

Answer:

Test scores of 10.2 or lower are significantly low.

Test scores of 31.4 or higher are significantly high.

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 = 20.8, \sigma = 5.3

Identify the test scores that are significantly low or significantly high.

Significantly low

Z = -2 and lower.

So the significantly low scores are thoses values that are lower or equal than X when Z = -2. So

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

-2 = \frac{X - 20.8}{5.3}

X - 20.8 = -2*5.3

X = 10.2

Test scores of 10.2 or lower are significantly low.

Significantly high

Z = 2 and higher.

So the significantly high scores are thoses values that are higherr or equal than X when Z = 2. So

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

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X - 20.8 = 2*5.3

X = 31.4

Test scores of 31.4 or higher are significantly high.

3 0
4 years ago
A bag of marbles contains 12 red marbles 8 blue marbles and 5 green marbles. If three marbles are pulled out find each of the pr
worty [1.4K]

Answer:  The probability of pulling a red marble is 48%, probability of pulling a blue marble is 32%, probability of pulling a green marble is 20%

and

the probability of pulling three green marbles out with replacement is 0.8%.

Step-by-step explanation:  Given that a bag of marbles contains 12 red marbles 8 blue marbles and 5 green marbles.

Let S be the sample space for the experiment of pulling a marble.

Then, n(S) = 12 + 8 + 5 = 25.

Let, E, F and G represents the events of pulling a red marble, a blue marble and a green marble respectively.

The, n(E) = 12, n(F) = 8  and  n(G) = 5.

Therefore, the probabilities of each of these three events E, F and G will be

P(E)=\dfrac{n(E)}{n(S)}=\dfrac{12}{25}=\dfrac{12}{25}\times100\%=48\%,\\\\\\P(F)=\dfrac{n(F)}{n(S)}=\dfrac{8}{25}=\dfrac{8}{25}\times100\%=32\%,\\\\\\P(G)=\dfrac{n(G)}{n(S)}=\dfrac{5}{25}=\dfrac{5}{25}\times100\%=20\%.

Now, the probability of pulling three green marbles out with replacement is given by

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Thus, the probability of pulling a red marble is 48%, probability of pulling a blue marble is 32%, probability of pulling a green marble is 20%

and

the probability of pulling three green marbles out with replacement is 0.8%.

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Answer:

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5 0
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