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3 years ago

Howard bought 3/5 pounds of granola. He divided the granola evenly among 4 friends. How much granola did each friend get?

Mathematics

1 answer:

Alex17521 [72]3 years ago

6 0

Answer:

(0.75) is the amount that each freind has.

Step-by-step explanation:

We see that Howard did NOT purchase not one pound of granola. Its only 60% of a pound. Therefore we'd do 60% divided by 4 and therefore get .75%. Each freind ate .75% of that granola.

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What is the probability that the sum is a multiple of 4 when you roll to number cubes

GREYUIT [131]

Answer:

Step-by-step explanation:

1/4

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3 years ago

Write the equation for the function graphed below.

avanturin [10]

Given the vertex of the parabola at point (2, -3):

The quadratic function in vertex form is:

f(x) = (x - 2)^2 - 3

where the vertex (h, k) is the minimum point = (2, -3).

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3 years ago

A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the

Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average age = 25 years

$\sigma$ = population standard deviation = 1.8 years

n = sample of students = 36

$\mu$ = population average age

So, 98% confidence interval for the average age, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

P( $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ]

= [ $25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }$ , $25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }$ ]

= [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0

3 years ago

Simplify:<br><br> {(-8)^-4 ÷ 2^-8}^2

Aleksandr-060686 [28]

Answer:

ok so first we have to do whats in the brackets then we have to do to exponits so first

(-0.00024414062divided by 0.00390625)^2

-0.06249999872^2

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Hope This Helps!!!

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