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2 years ago

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Donna has 4 3/4 cups of sugar. She plans to make muffins that take 1 2/3 cups of sugar per pan. Does she have enough to make 2 p

ans of muffins? Defend your answer.

1 answer:

maw [93]2 years ago

4 0

The answer is 2 and one half cups is If I’m correct and

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I need help as soon as possible someone please help

Zigmanuir [339]

Answer:

b

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

in a class of 40 students, 24 are over the age of 21 and 16 are 21 or younger. Biff is over the age of 21. A student is selected

LenKa [72]

Answer:

3. 60% probability the student is over the age of 21.4

4. 4.17% probability Biff was selected if you know the student selected was over the age of 21.

5. 0% probability Biff was selected if you know the student selected was 21 or younger

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcoes.

In this question:

40 students

24 over the age of 21.

16 are 21 or younger.

3. Find the probability the student is over the age of 21.

24 of the 40 students are over the age of 21.

24/40 = 0.6

60% probability the student is over the age of 21.4

4. Find the probability Biff was selected if you know the student selected was over the age of 21.

24 students over the age of 21, one of which is Biff.

1/24 = 0.0417

4.17% probability Biff was selected if you know the student selected was over the age of 21.

5. Find the probability Biff was selected if you know the student selected was 21 or younger

Biff is not 21 or younger, he is over 21, so there are no desired outcomes.

So, 0% probability Biff was selected if you know the student selected was 21 or younger

8 0

3 years ago

How can I find the prime factorization of a number with an easy explanation?

mezya [45]

Answer:

Step-by-step explanation:

Step 1: Start by dividing the number by the first prime number 2 and continue dividing by 2 until you get a decimal or remainder. Then divide by 3, 5, 7, etc. until the only numbers left are prime numbers. Step 2: Write the number as a product of prime numbers.

4 0

2 years ago

Help please its URGENT!!!!!!!!

Aneli [31]

This is a relation but not a function.

Functions must have that there is a unique y for a given x, which clearly doesn't work here because all lines of a given x have 2 y-values. However, it is a relation because there is a given set of points which are defined to be within the set of the ellipse (if it's defined which members of two sets, the range and domain, go together, then you have a relation)

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3 years ago

Read 2 more answers

A survey was conducted that asked 1003 people how many books they had read in the past year. Results indicated that x overbar eq

Sergio [31]

Answer:

The 99% confidence interval would be given (11.448;14.152).

Step-by-step explanation:

1) Important concepts and notation

A confidence interval for a mean "gives us a range of plausible values for the population mean. If a confidence interval does not include a particular value, we can say that it is not likely that the particular value is the true population mean"

$s=16.6$ represent the sample deviation

$\bar X=12.8$ represent the sample mean

n =1003 is the sample size selected

Confidence =99% or 0.99

$\alpha=1-0.99=0.01$ represent the significance level.

2) Solution to the problem

The confidence interval for the mean would be given by this formula

$\bar X \pm z_{\alpha/2} \frac{s}{\sqrt{n}}$

We can use a z quantile instead of t since the sample size is large enough.

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$12.8 - 2.58 \frac{16.6}{\sqrt{1003}}=11.448$

$12.8 + 2.58 \frac{16.6}{\sqrt{1003}} =14.152$

And the 99% confidence interval would be given (11.448;14.152).

We are confident that about 11 to 14 are the number of books that the people had read on the last year on average, at 1% of significance.

3 0

3 years ago