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3 years ago

Help!!!!!!!!!!!!!!!!

Mathematics

1 answer:

valkas [14]3 years ago

5 0

Under box 36x + 9, Drag the following:

A.) 9(4x + 1)
D.) (9 . 4x) + (9 . 1)

Under box 9(4x - 1), Drag the following:
B.) (3 . 12x) - (3 . 3)
C.) 36x - 9

Under box (4 . 9x) + (4 . 2), Drag the following:
E.) 4(9x + 2)
F.) 36x + 8

Hope this helps!

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A dogs weighs 7 pounds and triples each year for two years. How much does the dog weigh?

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Out of the 40 students in the 7th grade class, 60% of the students carry a cell phone. What fraction of the students in the 7th

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Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we

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Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N( $\mu = 7 lbs , \sigma^{2} = 0.875^{2} lbs$ )

The standard normal z distribution is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, X bar = sample mean weight

n = sample size = 4

Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)

P(X bar > 7.5) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{7.5-7}{\frac{0.875}{\sqrt{4} } }$ ) = P(Z > 1.1428) = 0.1283 (using z% table)

Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .

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3 years ago