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lutik1710 [3]
3 years ago
13

Polly buys 14 cupcakes for a party. The bakery puts them into boxes that hold 4 cupcakes each.

Mathematics
1 answer:
Temka [501]3 years ago
8 0

Answer:

Part a) The number of boxes needed will be 4

part b) The fraction of the box that is empty is \frac{1}{2}

Part c) Are needed 2 cupcakes to fill the last box

Step-by-step explanation:

Part a) How many boxes will be needed for Polly to bring all the cupcakes to the party?

To find out the number of boxes needed, divide the total number of cupcakes by 4 (maximum number of cupcakes that fit in each box).

so

\frac{14}{4}=3.5\ box

Round up

The number of boxes needed will be 4

Part b) If the bakery completely fills as many boxes as possible, what fraction of the last box is empty?

we know that

The maximum number of cupcakes per box is 4

The first three boxes have 4 cupcakes

The last box has

14-3(4)=2\ cupcakes

The fraction of the box that is empty is equal to the number of cupcakes missing to fill the box divided by the cupcake capacity of the box

\frac{2}{4}=\frac{1}{2}

Part c) How many more cupcakes are needed to fill this box

Subtract the number of cupcakes in the box from the cupcake capacity of the box

The last box has 2 cupcakes

The cupcake capacity of the box is 4

so

4-2=2\ cupcakes

therefore

Are needed 2 cupcakes to fill the last box

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Step-by-step explanation:

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3 0
3 years ago
12. If a manufacturer conducted a survey among randomly selected target market households and wanted to be 95% confident that th
atroni [7]

Answer:

n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11  

And rounded up we have that n=1068

Step-by-step explanation:

We have the following info given:

Confidence= 0.95 the confidence level desired

ME =0.03 represent the margin of error desired

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

The confidence level is 95% or 0.95, the significance is \alpha=0.05 and the critical value for this case using the normal standard distribution would be z_{\alpha/2}=1.96

Since we don't have prior information we can use \hat p= 0.5 as an unbiased estimator

Also we know that ME =\pm 0.03 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11  

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6 0
3 years ago
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Answer:

\boxed{ \bold{ \huge{ \boxed{ \sf{1 \frac{2}{15}}}}}}

Step-by-step explanation:

\sf{1 \frac{7}{10}  \times  \frac{2}{3} }

Convert the mixed number into improper fraction

\longrightarrow{ \sf{ \frac{10 \times 1 + 7}{10}  \times  \frac{2}{3}}}

\longrightarrow{ \sf{ \frac{17}{10}  \times  \frac{2}{3}}}

To multiply one fraction by another, multiply the numerators for the numerator, and multiply the denominator for its denominator and reduce the fraction obtained after multiplication into lowest term

\longrightarrow{ \sf{ \frac{17 \times 2}{10 \times 3}}}

\longrightarrow{ \sf{ \frac{34}{30}}}

\longrightarrow{ \sf{ \frac{17}{15} }}

Convert the improper fraction into mixed number

\longrightarrow{ \sf{1 \frac{2}{15} }}

Hope I helped!

Best regards! :D

7 0
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Bezzdna [24]

Answer:

4 :45 pm

Step-by-step explanation:

2 : 20 + 17 minutes = 2:37

2:37 + 1hr 1 minute  = 3 : 38

3 : 38 + 1 hr 7 = 4 :45 pm

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