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kozerog [31]
3 years ago
9

Can someone help? TvT please

Mathematics
1 answer:
zysi [14]3 years ago
7 0

Answer:

I don't know bro actually which question ❓

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Use deductive reasoning to show that the following procedure always produces the number 6. Procedure: Pick a number. Add 4 to th
Flauer [41]

Answer:

procedure always produces 6

Step-by-step explanation:

Let 'n' be the unknown number

Add 4 to the number : n+4

multiply the sum by 3.

multiply the sum n+4 by 3

3(n+4) is 3n+12

Now subtract 6, so we subtract 6 from 3n+12

3n+12-6=3n+6

finally decrease the difference by the tripe of the original number

triple of original number is 3n

3n+6-3n= 6

so the procedure always produces 6

5 0
3 years ago
A cube has a length, width and height of 6 feet. What is the volume in cubic INCHES?
rodikova [14]

Volume of cube = ( side)^3

Side = 6 feet = 6×12 = 72 inches ( 1ft =12 inch)

Vol. = (72)^3 = 72×72×72= 373248 cubic inches

6 0
3 years ago
X +4/2 = 9<br> What’s z?
kirill [66]

Answer:

x+4/2=9       x=14

Step-by-step explanation:

14+4=18 and 19/2 is equal to 9. Therefore the answer is 14.

6 0
3 years ago
QUICK HELP!!!<br> How manny differences can you spot? <br> 8<br> 6<br> 7<br> 4<br> 3<br> 5
Eduardwww [97]

Answer:

i can only see 3

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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
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