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professor190 [17]

3 years ago

15

two kilograms of ground cinnamon is packaged into bags containing 38 grams each.there will also be cinnamon left over.how many b

ags will there be?

1 answer:

Fantom [35]3 years ago

7 0

There will be 52 full bags plus an additional one that is not quite full yet.

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igomit [66]

0.03125 beciase there are 32 people and one ticket make that as a fraction which is 1/32 and make it a decimal

7 0

3 years ago

GIVING BRAINLIEST ONCE AGAIN

Andru [333]

Answer:

45

Step-by-step explanation:

divide 25 by 5 to see how they got that, which is 5. then multiply 9 by 5 bc you have to do the same thing, which is 45.

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3 years ago

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Bella types at a constant rate of 42 words per minute. Is the number of words she can type proportional to the number of minutes

solmaris [256]

Yes, the number of words ARE proportional to the number of minutes she types, away to tell is when it says Bella types at a CONSTANT rate of 42 words per minute. This means for every minute 42 words will be added.

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3 years ago

If a drink costs 2.32, how many drinks can you buy with 13 dollars?

Llana [10]

Answer:

5 drinks

Step-by-step explanation:

If the drinks cost $2.32 and you have $13 you can divide 13 by 2.32

13/2.32+ 5.6034...

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3 years ago

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Suppose that an airline uses a seat width of 16.5 in. Assume men have hip breadths that are normally distributed with a mean of

Alexxx [7]

Answer:

a) 0.018

b) 0

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 14.4 in

Standard Deviation, σ = 1 in

We are given that the distribution of breadths is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(breadth will be greater than 16.5 in)

P(x > 16.5)

$P( x > 16.5) = P( z > \displaystyle\frac{16.5 - 14.4}{1}) = P(z > 2.1)$

$= 1 - P(z \leq 2.1)$

Calculation the value from standard normal z table, we have,

$P(x > 16.5) = 1 - 0.982 = 0.018 = 1.8\%$

0.018 is the probability that if an individual man is randomly selected, his hip breadth will be greater than 16.5 in.

b) P( with 123 randomly selected men, these men have a mean hip breadth greater than 16.5 in)

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

P(x > 16.5)

$P( x > 16.5) = P( z > \displaystyle\frac{16.5-14.4}{\frac{1}{\sqrt{123}}}) = P(z > 23.29)$

$= 1 - P(z \leq 23.29)$

Calculation the value from standard normal z table, we have,

$P(x > 16.5) = 1 - 1 = 0$

There is 0 probability that 123 randomly selected men have a mean hip breadth greater than 16.5 in

4 0

3 years ago