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AveGali [126]
2 years ago
8

Emily and Kate were at a cheerleading competition this fall.They cheered 12 rounds. Each round lasted 8.24 minutes. Calculate th

e total number of minutes, Emily cheered and the total number of minutes Kate cheered at the competition
Mathematics
1 answer:
trapecia [35]2 years ago
4 0

Answer:

98.88

Step-by-step explanation:

Given: They cheered 12 rounds. Each round = 8.24

To find the total number, multiply how much one round is with the total number of rounds.

  8.24

×     12

-----------

  1648

+ 8240

----------

98.88

Each girl cheered 98.88 minutes. If you need this estimated, the answer would be 99 minutes.

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A would be true

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C would be false

D would be true, assuming "about" means you can round up a little.

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6 0
3 years ago
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Steven earns extra money babysitting. he charges $20.25 for 3 hours and $40.50 for 6 hours.
il63 [147K]
Y=6.75x
He charges $6.75 per hour
6 0
2 years ago
Is -2 a solution to 4x^(3)-x^(2)+16x-4
sergiy2304 [10]

Answer:

Not a solution

Step-by-step explanation:

In order for x=-2 to be a solution, the result after substituting must be 0:

4(-2)^3-(-2)^2+16(-2)-4\stackrel{?}{=}0

4(-8)-4-32-4\stackrel{?}{=}0

-32-40\stackrel{?}{=}0

-72\neq0

Therefore, x=-2 is not a solution to 4x^3-x^2+16x-4

3 0
2 years ago
A different children's party cost £225.50 How many children were at the party?​
Alecsey [184]
Hi, I think you forgot to write a part of the problem, could you write the rest in the comments please. I would be happy to help you
8 0
3 years ago
The tensile strength of a metal part is normally distributed with mean 40 pounds and standard deviation 5 pounds. If 50,000 part
nalin [4]

Answer:

a) how many would you expect to fail to meet a minimum specification limit of 35-pounds tensile strength?

7933 parts

b) How many would have a tensile strength in excess of 48 pounds?

2739.95 parts

Step-by-step explanation:

The formula for calculating a z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

a) how many would you expect to fail to meet a minimum specification limit of 35-pounds tensile strength?

z = (x-μ)/σ

x = 35 μ = 40 , σ = 5

z = 35 - 40/5

= -5/-5

= -1

Determining the Probability value from Z-Table:

P(x<35) = 0.15866

Converting to percentage = 15.866%

We are asked how many will fail to meet this specification

We have 50,000 parts

Hence,

15.866% of 50,000 parts will fail to meet the specification

= 15.866% of 50,000

= 7933 parts

Therefore, 7933 parts will fail to meet the specifications.

b) How many would have a tensile strength in excess of 48 pounds?

z = (x-μ)/σ

x = 48 μ = 40 , σ = 5

z = 48 - 40/5

z = 8/5

z = 1.6

P-value from Z-Table:

P(x<48) = 0.9452

P(x>48) = 1 - P(x<48)

1 - 0.9452

= 0.054799

Converting to percentage

= 5.4799%

Therefore, 5.4799% will have an excess of (or will be greater than) 48 pounds

We are asked, how many would have a tensile strength in excess of 48 pounds?

This would be 5.4799% of 50,000 parts

= 5.4799% × 50,000

= 2739.95

Therefore, 2739.95 parts will have a tensile strength excess of 48 pounds

4 0
3 years ago
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