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3 years ago

7

The speeders soccer team charged $12 to wash each car at a fundraiser car wash.The team collected a total os $672 by the end of

the day.How many cars did the team wash?

2 answers:

mash [69]3 years ago

8 0

the answer is 56 because 672 divided by 12 is 56 cars

sukhopar [10]3 years ago

4 0

They washed 56 cars because 672/12=56

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A team of three students are working on a language-learning app; they need to develop 300 micro-lessons and 300 micro-tests befo

Olegator [25]

Answer: a) 15 b)

Step-by-step explanation:

Let X be the number of days:

a)

For LESSONS:

Jordan does 10 / day ( 10*X)

Marco 5 / day ( 5*X)

Junyi 5 / day ( 5*X)

For TESTS:

Jordan does 5 / day ( 5*X)

Marco 10 / day ( 10*X)

Junyi 8 / day ( 8*X)

for each they need a total of 300

a) $10X+5X+5X=300 => 20X = 300 => X = 15$ days for the lessons

b) $5X+10X+8X = 300 => 23X = 300 => X = 13.04$ days for the tests

so they need 15 days to finish both tasks

now if Junyi gets sick we just eliminate his contribution

a) $10X+5X=300 => 15X = 300 => X = 20$ days for the lessons

b) $5X+10X = 300 => 15X = 300 => X = 20$ days for the tests

so in 20 days they will finish without him

If jordan works 10 hours a day, we just replace him with 10/24

a) $10(10/24)+5X+5X= 300 => X = 29.58$ days for the lessons

b) $5(10/24)+10X+8X = 300 => X = 16.51$ days for the tests

so at the end to complete both tasks they need 29.58 days

4 0

3 years ago

15 pts! Will Give Brainliest Halp! PLZ

aivan3 [116]

Theres a saying: Dividing fractions dont ask why, just flip the second and multiply.

so a/b / c/d = a/b * d/c

-8/2 * -3/6

1. cross - cancel common factor 2

4/2 * -3/3

2. Multiply

-4(-3)/ 2*3

3. multiply numbers

--12/2*3

- -12/6

4. Apply fraction rule -a/b=- a/b

=-(-12/6)

5. divide

= -(-2)

6. Apply rule -(-a)=a

=2

Your answer would be 2

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3 years ago

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What can be concluded about the sphere? Check all that apply. The sphere has a radius of 10 cm. The diameter measure is substitu

MA_775_DIABLO [31]

Answer: Its C and E on engenuity

Step-by-step explanation:

The radius is half the diameter.

The volume of the sphere is two-thirds the volume of a cylinder with the same radius and height.

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3 years ago

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Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean

Mamont248 [21]

Answer:

a) The percentage of snails that take more than 60 hours to finish is 4.75%.

b) The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c) The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d) 0% probability that a randomly-chosen snail will take more than 76 hours to finish

e) To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f) The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 50, \sigma = 6$

a. The percentage of snails that take more than 60 hours to finish is

This is 1 subtracted by the pvalue of Z when X = 60.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

1 - 0.9525 = 0.0475

The percentage of snails that take more than 60 hours to finish is 4.75%.

b. The relative frequency of snails that take less than 60 hours to finish is

This is the pvalue of Z when X = 60.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c. The proportion of snails that take between 60 and 67 hours to finish is

This is the pvalue of Z when X = 67 subtracted by the pvalue of Z when X = 60.

X = 67

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{67 - 50}{6}$

$Z = 2.83$

$Z = 2.83$ has a pvalue 0.9977

X = 60

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

0.9977 - 0.9525 = 0.0452

The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d. The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 1 subtracted by the pvalue of Z when X = 76.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{76 - 50}{6}$

$Z = 4.33$

$Z = 4.33$ has a pvalue of 1

1 - 1 = 0

0% probability that a randomly-chosen snail will take more than 76 hours to finish

e. To be among the 10% fastest snails, a snail must finish in at most hours.

At most the 10th percentile, which is the value of X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f. The most typical 80% of snails take between and hours to finish.

From the 50 - 80/2 = 10th percentile to the 50 + 80/2 = 90th percentile.

10th percentile

value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

90th percentile.

value of X when Z has a pvalue of 0.9. So X when Z = 1.28

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 50}{6}$

$X - 50 = 1.28*6$

$X = 57.68$

The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

5 0

3 years ago

Find the volume of the composite figure

natulia [17]

I dont know maybe I just want points
2

8 0

3 years ago