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Serga [27]
3 years ago
6

Camillo needs 2,400 oz of jelly for the food challenge. If 48 oz of jelly cost $3.84, how much will Camillo spend on jelly? Expl

ain how you can find your answer. brainliest is on the table
Mathematics
1 answer:
olga2289 [7]3 years ago
3 0

My Answer:

Well he will spend I a lot I will say. But the steps to get to find out the answer are not that hard. The first step is to find out how much he needs. For this challenge he need 2,400 oz of jelly. Next we will divide that number by 48 oz. That number is how many jellies you will get if you spend $3.84. When you divide that you will get 50. Now you will multiply 50 and 3.84. After you multiply that you will get $192. That is how much Camillo is spending on jellies. I hope this has answerd your question

Another possible answer:

First, find the unit price of the jelly. The unit cost of jelly is $0.08 per ounce. Next, find the total price of 2,400 oz by multiplying the unit price by the quantity. The total price is $192.

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Based on the information given, the temperature when converted to Fahrenheit will be 104° Fahrenheit.

<h3>Solving the temperature.</h3>

It should be noted that when converting to Fahrenheit from Celcius, one has to multiply by 1.8 and then add to 32.

In this case, the temperature given in Celcius is 40°. Therefore, the Fahrenheit will be;

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When graphing 5x-3y&gt;30 do we shade above or below?
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In a simple random sample of 300 boards from this shipment, 12 fall outside these specifications. Calculate the lower confidence
Lyrx [107]

Answer:

The 95% confidence interval for the percentage of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

Step-by-step explanation:

In a random sample of 300 boards the number of boards that fall outside the specification is 12.

Compute the sample proportion of boards that fall outside the specification in this sample as follows:

\hat p =\frac{12}{300}=0.04

The (1 - <em>α</em>)% confidence interval for population proportion <em>p</em> is:

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

The critical value of <em>z</em> for 95% confidence level is,

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

*Use a <em>z</em>-table.

Compute the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification as follows:

CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.04\pm1.96\sqrt{\frac{0.04(1-0.04)}{300}}\\=0.04\pm0.022\\=(0.018, 0.062)\\\approx(1.8\%, 6.2\%)

Thus, the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

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