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coldgirl [10]
3 years ago
12

A cake recipe calls for 2 3/4 cups of sugar. A caterer has 14 1/2 cups of sugar on hand. How many cakes can he make?(answer need

s to be rounded down to a whole number)
Mathematics
1 answer:
Mariana [72]3 years ago
3 0

Answer:

5 Cakes

Step-by-step explanation:

14.5/2.75=5.27

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What is th rate equivalent to an annual rate of 20% with monthly capitalization​
Strike441 [17]

Answer:

  about 21.9%

Step-by-step explanation:

The effective annual multiplier is ...

  (1 +r/n)^n = (1 +0.20/12)^12 ≈ 1.21939

The rate equivalent is 1 less than this:

  1.21939 -1 = 0.21939 = 21.939%

6 0
3 years ago
A soft-drink machine at a steakhouse is regulated so that the amount of drink dispensed is normally distributed with a mean of 2
AnnZ [28]

Answer:

z=\frac{203-200}{\frac{15}{\sqrt{9}}}=0.6    

p_v =P(z>0.6)=0.274  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can't conclude that the true mean is not significantly higher than 200 at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=203 represent the mean height for the sample  

\sigma=15 represent the population standard deviation

n=9 sample size  

\mu_o =200 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 200, the system of hypothesis would be:  

Null hypothesis:\mu \leq 200  

Alternative hypothesis:\mu > 20  

If we analyze the size for the sample is < 30 but we  know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

z=\frac{203-200}{\frac{15}{\sqrt{9}}}=0.6    

P-value

Since is a one sided test the p value would be:  

p_v =P(z>0.6)=0.274  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can't conclude that the true mean is not significantly higher than 200 at 5% of signficance.  

8 0
3 years ago
4-30. Micron Technology has sales offices located in four
ivanzaharov [21]

Answer:

An invoice was paid an average of 20 days after it was received.

Mean =

Standard deviation =

Now we are supposed to find what percent of the invoices were paid within 15 days of receipt i.e.P(x<15)

Formula :

At x = 15

Substitute the values

Refer the z table for p value

So, p value = 0.1587

So, 15.87% of the invoices were paid within 15 days of receipt

Step-by-step explanation:

6 0
3 years ago
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