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3 years ago

Sherri bought 45 cupcakes for a party she was having. If 60% of the cupcakes, were chocolate, how many cupcakes were chocolate?

Mathematics

2 answers:

Veseljchak [2.6K]3 years ago

4 0

27 cupcakes were chocolate

djyliett [7]3 years ago

3 0

27 cupcakes were chocolate

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A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient

ankoles [38]

Answer:

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

Solution:

We need to show that the gradient of the curve at A is 1

Here given that ,

$y=(x-a) \sqrt{(x-b)}$ --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

$0=(b+1-a) \sqrt{(b+1-b)}$

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

$y^{\prime}=u^{\prime} y+y^{\prime} u$

so, we get

${u}^{\prime}=1$

$v^{\prime}=\frac{1}{2} \sqrt{(x-b)}$

$y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}$

By putting value of point A and putting value of eq 2 we get

$y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}$

$y^{\prime}=\frac{d y}{d x}=1$

Hence proved that the gradient of the curve at A is 1.

7 0

2 years ago

Find the sum of 37, 9, 663, 1198, and 45 a. 1952 b. 1142 c. 942 d. 3952 e. 2722

frez [133]

The answer is 1952 because you have to add all of the numbers together

5 0

2 years ago

Read 2 more answers

Logan spends 26% of his weekly allowance on 2 banana splits. Banana splits cost $3.25 each at Frozen Treats. How much money does

Shtirlitz [24]

I am not sure u should google it plz

6 0

3 years ago

A savings and loan association needs information concerning the checking account balances of its local customers. A random sampl

Marina86 [1]

Answer:

The 99% confidence interval for the true mean checking account balance for local customers is ($439.29, $888.99).

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 14 - 1 = 13

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 13 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$ . So we have T = 3.0123

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 3.0123\frac{279.29}{\sqrt{14}} = 224.85$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 664.14 - 224.85 = $439.29

The upper end of the interval is the sample mean added to M. So it is 664.14 + 224.85 = $888.99.

The 99% confidence interval for the true mean checking account balance for local customers is ($439.29, $888.99).

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3 years ago

A factory inspector found flaws in 3 out of 18 wooden boxes what is the experimental probability that the next wooden box will b

Sphinxa [80]

Answer:

16.666% would be flawed.

Step-by-step explanation:

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2 years ago