All categories

3 years ago

There are 105 calories in a glass and a half of lemonade. how many calories are in 2 1/2 glasses

Mathematics

2 answers:

ddd [48]3 years ago

8 0

Answer:

175

Step-by-step explanation:

A glass and a half is 1+1/2 = 3/2 glasses.

2 1/2 glasses is 2/2 glasses more than that, or 5/2 glasses.

5/2 glasses is 5/3 times the size of 3/2 glasses, so there will be 5/3 the number of calories:

... (105 calories) × (5/3) = 175 calories

lbvjy [14]3 years ago

5 0

Answer:

The answer is 175 calories in 2 1/2 glasses

Step-by-step explanation:

x = 2.5 x 105/1.5

x = 262.5/1.5

x = 175

You might be interested in

Given the speeds of each runner below, determine who runs the fastest.

Lesechka [4]

Answer:

Debbie runs 9 feet per second: 9 ft/s

Jessica runs 1 mile in 440 seconds: 414/50 ft/s = 207/25 ft/s = 8 7/25 ft/s

jesssica runs 131 feet in 10 seconds: since 1 mile = 5280 ft we have 1/470 mi/s = 5280/470 ft/s = 528/47 ft/s = 11 11/47 ft/s

ron runs 603 feet in 1 minute: 547 ft/min = 603/60 ft/s = 201/20 ft/s = 10 1/20 ft/s

Since:

11 11/47 > 10 1/20 > 9 > 8 7/25

Emily runs the fastest.

7 0

3 years ago

Lily is a botanist who works for a garden that many tourists visit. The function f(s) = 2s + 30 represents the number of flowers

Vesnalui [34]

Answer:

A. b(w) = 80w +30

B. input: weeks; output: flowers that bloomed

C. 2830

Step-by-step explanation:

For f(s) = 2s +30, and s(w) = 40w, the composite function f(s(w)) is ...

b(w) = f(s(w)) = 2(40w) +30

b(w) = 80w +30 . . . . . . blooms over w weeks

The input units of f(s) are <em>seeds</em>. The output units are <em>flowers</em>.

The input units of s(w) are <em>weeks</em>. The output units are <em>seeds</em>.

Then the function b(w) above has input units of <em>weeks</em>, and output units of <em>flowers</em> (blooms).

For 35 weeks, the number of flowers that bloomed is ...

b(35) = 80(35) +30 = 2830 . . . . flowers bloomed over 35 weeks

7 0

2 years ago

Which of the following is the inverse of F(x) if F(x) = 6x?

exis [7]

The best answer is d

8 0

3 years ago

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as t

skad [1K]

Answer:

a) 0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

b) 0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

c) 0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

d) None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as the population mean and assume the population standard deviation of preparation fees is $100.

This means that $\mu = 273, \sigma = 100$

A) What is the probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 30, s = \frac{100}{\sqrt{30}}$

The probability is the p-value of Z when X = 273 + 16 = 289 subtracted by the p-value of Z when X = 273 - 16 = 257. So

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{30}}}$

$Z = 0.88$

$Z = 0.88$ has a p-value of 0.8106

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{30}}}$

$Z = -0.88$

$Z = -0.88$ has a p-value of 0.1894

0.8106 - 0.1894 = 0.6212

0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

B) What is the probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 50, s = \frac{100}{\sqrt{50}}$

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{50}}}$

$Z = 1.13$

$Z = 1.13$ has a p-value of 0.8708

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{50}}}$

$Z = -1.13$

$Z = -1.13$ has a p-value of 0.1292

0.8708 - 0.1292 = 0.7416

0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

C) What is the probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 100, s = \frac{100}{\sqrt{100}}$

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{100}}}$

$Z = 1.6$

$Z = 1.6$ has a p-value of 0.9452

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{100}}}$

$Z = -1.6$

$Z = -1.6$ has a p-value of 0.0648

0.9452 - 0.0648 =

0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

D) Which, if any of the sample sizes in part (a), (b), and (c) would you recommend to ensure at least a .95 probability that the same mean is withing $16 of the population mean?

None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

6 0

2 years ago

(Use distributive property)<br> 6 (5x + 3)

S_A_V [24]

Answer:

30x + 18

Step-by-step explanation:

Using distributive property, you can find the answer like so.

Multiply 6 and 5x first

6(5x + 3)

30x + 3

Now multiply 6 by 3

6(5x + 3)

30x + 18

Hope I explained this well to you. :)

7 0

2 years ago