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2 years ago

PLEASE PLEASE PLEASE PLEASE (5'1, 5'2, 5'3, 5'4, 5'5, 5'6, 5'7, 5'8, 5'9, 5'9, 5'10, 5'11, 5'12)

Mathematics

1 answer:

Yakvenalex [24]2 years ago

4 0

Answer:5

5 feet 7 inches

Step-by-step explanation:

first add all of the given together and then divide what you get by the number of terms given

(5'2+5'3+5'4+ 5'5+5'6+ 5'7+ 5'8+5'9+5'9,+5'10+5'11+5'12)/12 = 5'7 11/64 inches

then divide 11 by 64 on a calculator and then round the decimal you get to the nearest tenth which gives you the answer 5 feet 7 inches.

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PLEASE HELP! IF YOU DON'T KNOW DON'T ANSWER! 60 POINTS!!

postnew [5]

Answer:

an = 1/2 (4)^ (n-1)

a6 = 512

Step-by-step explanation:

The formula for a geometric sequence is

an = a1 (r)^(n-1)

where an is the term of the sequence

a1 is the initial term of the sequence

r is the ratio

and n is the term number

We know a1 = 1/2 and r =4

I will assume that x=6 means we want to know the 6th term

an = 1/2 (4)^ (n-1)

We want to find the 6th term

a6 = 1/2 * 4^(6-1)

a6 = 1/2 * 4^5

a6 = 512

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3 years ago

Planet Krypton has 2 times the radius and 2 times the mass of the Earth. What is the best estimate for the magnitude of the grav

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Answer:

Step-by-step explanation:

it's a because it would be ×2

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3 years ago

HELP ME I was out of school when we learned this from a concussion and broken arm

tresset_1 [31]

Answer:

2 flavors and 4 toppings

Step-by-step explanation:

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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

Elodia [21]

Answer:

7.3% percentage of the bearings produced will not be acceptable.

Step-by-step explanation:

Consider the provided information.

Average diameter of the bearings it produces is .500 inches. A bearing is acceptable if its diameter is within .004 inches of this target value.

Let X is the normal random variable which represents the diameter of bearing.

Thus, 0.500-0.004<X<0.500+0.004

0.496<X<0.504

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Use the Z score formula: $\frac{X-\mu}{\sigma}$

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$\frac{0.496-0.499}{0.002}\leq z\leq \frac{0.504-0.499}{0.002}$

$\frac{-0.003}{0.002}\leq z\leq \frac{0.005}{0.002}$

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Now use the standard normal table and determine the probability of that a ball bearing will be acceptable.

$P(-1.5\leq z\leq 2.5)=0.9938-0.0668=0.9270$

We need to find the percentage of the bearings produced will not be acceptable.

So subtract it from 1 as shown.

1-0.9270=0.073

Hence, 7.3% percentage of the bearings produced will not be acceptable.

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The line g(x) is 3 units above the line f(x). k = 3.

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