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AURORKA [14]
1 year ago
5

Tina baked a batch of 40 cupcakes. daisy baked 4/5 as many cupcakes as tina. miko baked 1/2 as many cupcakes as daisy. how many

cupcakes did the 3 girls bake altogether?
Mathematics
1 answer:
Alex17521 [72]1 year ago
6 0

The three girls baked 88 cupcakes altogether.

Cupcakes Baked by Each

Number of cupcakes baked by Tina, T = 40

It is given that Daisy baked 4/5 as many cupcakes as Tina.

Therefore, number of cupcakes baked by Daisy, D = (4T/5) (Applying fractions)

⇒ D = (4/5)×40

⇒ D = 4 × 8

⇒ D = 32

Also, it is given that Miko baked 1/2 as many cupcakes as Daisy.

Thus, number of cupcakes baked by Miko, M = D/2 (Applying fractions)

⇒  M = 32/2

⇒ M = 16

Calculating Total Number of Cupcakes Baked

Cupcakes baked by the three girls altogether = T + D + M

= 40 + 32 + 16

= 88

Hence, the girls together baked 88 cupcakes.

Learn more on fractions here:

brainly.com/question/10354322

#SPJ4

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a) n=\frac{0.76(1-0.76)}{(\frac{0.01}{1.64})^2}=4905.83  

And rounded up we have that n=4906

b) For this case since we don't have prior info we need to use as estimatro for the proportion \hat p =0.5

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And rounded up we have that n=6724

Step-by-step explanation:

We need to remember that the confidence interval for the true proportion is given by :  

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

Part a

The estimated proportion for this case is \hat p =0.76

Our interval is at 90% of confidence, and the significance level is given by \alpha=1-0.90=0.1 and \alpha/2 =0.05. The critical values for this case are:

z_{\alpha/2}=-1.64, t_{1-\alpha/2}=1.64

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 margin of error desired is given ME =\pm 0.01 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)  

Replacing we got:

n=\frac{0.76(1-0.76)}{(\frac{0.01}{1.64})^2}=4905.83  

And rounded up we have that n=4906

Part b

For this case since we don't have prior info we need to use as estimatro for the proportion \hat p =0.5

n=\frac{0.5(1-0.5)}{(\frac{0.01}{1.64})^2}=6724  

And rounded up we have that n=6724

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3 years ago
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