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Vsevolod [243]
2 years ago
11

Cary's barbecue sauce recipe calls for 2 2/3 cups of water for every 8 pints of tomato sauce. How many cups of water does cary u

se when she makes barbecue sauce with 1 pint of tomato sauce
Mathematics
2 answers:
harina [27]2 years ago
7 0
In order to make this easier, let us convert the fraction into a mixed number. 2 2/3 would become 8/3. Now, let us divide 8 pints by 8/3 cups and we get 24/8 or 3 cups. Therefore, the amount of water that Cary has to use when she makes barbecue sauce with 1 pint of tomato is 3 cups. 
podryga [215]2 years ago
4 0

Answer:

1/3 cup of water

Step-by-step explanation:

The number of cups of water varies directly as the number of pints of tomato sauce. It means that as the number of cups of water used increases, so does the number of the pints of tomato sauce and vice versa.

Given that the barbecue sauce recipe calls for 2 2/3 cups of water for every 8 pints of tomato sauce, then 1 pint of tomato sauce would require an equivalent number T of water cups where

T = 1/8 *2 2/3

= 1/8 * 8/3

= 1/3

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Suppose that 80 percent of all statisticians are shy, whereas only 15 percent of all economists are shy. Suppose also that 90 pe
Anna71 [15]

Answer:

37.21% probability that the person is a statistician.

Step-by-step explanation:

We have these following probabilities:

An 80% probability that a statistican is shy.

A 15% probability that an economist is shy.

At the gathering, a 90% probability that a person is an economist.

At the gathering, a 10% probability that a person is a statistican.

If you meet a shy person at random at the gathering. What is the probability that the person is a statistician?

This can be formulated as the following question:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

P = \frac{P(B).P(A/B)}{P(A)}

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

So

What is the probability that the person is a statistican, given that he is shy?

P(B) is the probability of the person being a statistican. So P(B) = 0.1.

P(A/B) is the probability of a statistican being shy, so P(A/B) = 0.8.

P(A) is the probability of a person being a shy. This is 15% of 90%(economists) and 80% of 10%(statisticans). So

P(A) = 0.15*0.9 + 0.8*0.1 = 0.215

Finally

P = \frac{P(B).P(A/B)}{P(A)}

P = \frac{0.1*0.8}{0.215} = 0.3721

37.21% probability that the person is a statistician.

3 0
2 years ago
On the last day of vacation, the Ramirez family drove 252 miles from a state park to its home. The drive covered 35% of the tota
kirill115 [55]

Answer:

Step-by-step explanation:

35 percent of total miles traveled (say T)is 252 miles.

(35/100) * T=252

T=(252*100)/35 => T=720 miles

So total distance the Ramirez family traveled during its vacation is 720 miles

4 0
2 years ago
Read 2 more answers
I see that your new product is available for 412.50 on the website. How much is it if I buy it in the store? The website is a 25
OverLord2011 [107]

Answer with explanation:

Price of Product on the Website = $ 412.50

It is given that, product is offered at a discount of 25% on the website.

Let actual price of product on the store = $ x

Writing the above statement in terms of equation

→Price at store - Discount= Price at Website

x-\frac{25\times x}{100}=412.50\\\\ \frac{75 x}{100}=412.50\\\\x=\frac{41250}{75}\\\\x=550

Price at store of that Product = $ 550

3 0
2 years ago
Which equations represent a line that passes through the points given in the table? Check all that apply.
solong [7]

Answer:

I believe the answer is the 2nd box{y-2=1/6 (x+10)}, 3rd box{y-1=1/6(x+4)}, and the 5th box{y=1 x/6 +1/3}.

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
The lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.512.512, point, 5 years; the stan
zaharov [31]

This question was not written properly

Complete Question

The lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.5 years; the standard deviation is 2.4 years. Use the empirical rule (68-95-99.7\%)(68−95−99.7%) to estimate the probability of a lion living between 5.3 to 10. 1 years.

Answer:

Thehe probability of a lion living between 5.3 to 10. 1 years is 0.1585

Step-by-step explanation:

The empirical rule formula states that:

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.

3) 99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.

Mean is given in the question as: 12.5

Standard deviation : 2.4 years

We start by applying the first rule

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.

μ - σ

12.5 -2.4

= 10.1

We apply the second rule

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.

μ – 2σ

12.5 - 2 × 2.4

12.5 - 4.8

= 7.7

We apply the third rule

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.

μ - 3σ

= 12.5 - 3(2.4)

= 12.5 - 7.2

= 5.3

From the above calculation , we can see that

5.3 years corresponds to one side of 99.7%

Hence,

100 - 99.7%/2 = 0.3%/2

= 0.15%

And 10.1 years corresponds to one side of 68%

Hence

100 - 68%/2 = 32%/2 = 16%

So,the percentage of a lion living between 5.3 to 10. 1 years is calculated as 16% - 0.15%

= 15.85%

Therefore, the probability of a lion living between 5.3 to 10. 1 years

is converted to decimal =

= 15.85/ 100

= 0.1585

8 0
2 years ago
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