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2 years ago

What is 15 times 3 by 7

Mathematics

2 answers:

valentinak56 [21]2 years ago

7 0

Answer:

6.42857142857

Step-by-step explanation:

Stolb23 [73]2 years ago

4 0

Answer:

6 3/7

Step-by-step explanation:

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Rewrite in simplest terms -5(3u + 1) + 10

Luda [366]

Answer:

-15u+5

Step-by-step explanation:

open up the bracket -15u-5+10
liketerms together -15u+5
hence answer is -15u+5

8 0

2 years ago

Which equation demonstrates the associative property of multiplication

kherson [118]

Answer:

from what I remember ide say A

3 0

3 years ago

A random sample has been taken from a population. A statistician, using this sample, needs to decide whether to construct a 90 p

AURORKA [14]

Answer:

Step-by-step explanation: a confidence interval is a range that shows how precise the measurement is. if a statistician decides to construct using 90 percent C I over a 95% C I, it tells you that you have 10% chance of being wrong rather than 5% chance of being wrong while using the 95%. Also a 90 percent confidence interval would be narrower than a 95 percent confidence interval

5 0

2 years ago

The worlds two highest dams are both in Tajikistan. The rogue dam is 35 meters taller than the Nurek dam. together they are 635

galina1969 [7]

This is a sum and difference problem, with the equations
R-N=35
R+N=635
The solution of which is
R=(635+35)/2=335
N=(635-35)/2=300

In general, the sum and difference problem can be solved (most of the time mentally) using
Larger number = (sum + difference)/2
Smaller number = (sum - difference)/2

5 0

3 years ago

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

Marina86 [1]

Answer:

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 .

8 0

3 years ago