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2 years ago

Juan baked 250 cookies in 2 hours, How many cookies can he make in 1 minute?

Mathematics

1 answer:

dybincka [34]2 years ago

8 0

Answer:

I would say 2 because the decimal I came out with was 2.09

Step-by-step explanation:

You might be interested in

Plz help it’s due today

lina2011 [118]

Answer:

From left to right:

1/8

5/8

1-3/16

1-7/8

2-1/2

3-1/8

4-3/8

5-3/8

5-7/8

8 0

3 years ago

According to the American Lung Association, 7% of the population has lung disease. Of those having lung disease, 90% are smokers

postnew [5]

Answer:

Probability that a smoker has lung disease = 0.2132

Step-by-step explanation:

Let L = event that % of population having lung disease, P(L) = 0.07

So,% of population not having lung disease, P(L') = 1 - P(L) = 1 - 0.07 = 0.93

S = event that person is smoker

% of population that are smokers given they are having lung disease, P(S/L) = 0.90

% of population that are smokers given they are not having lung disease, P(S/L') = 0.25

We know that, conditional probability formula is given by;

P(S/L) = $\frac{P(S\bigcap L)}{P(L)}$

$P(S\bigcap L)$ = P(S/L) * P(L)

= 0.90 * 0.07 = 0.063

So, $P(S\bigcap L)$ = 0.063 .

Now, probability that a smoker has lung disease is given by = P(L/S)

P(L/S) = $\frac{P(S\bigcap L)}{P(S)}$

P(S) = P(S/L) * P(L) + P(S/L') * P(L')

= 0.90 * 0.07 + 0.25 * 0.93 = 0.2955

Therefore, P(L/S) = $\frac{0.063}{0.2955}$ = 0.2132

Hence, probability that a smoker has lung disease is 0.2132 .

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cint%5Climits%5E4_3%7Bx%5E%7B2%7D-x%20%7D%20%5C%2C%20dx" id="TexFormula1" title="\int\limits

melisa1 [442]

Answer:

$\frac{53}{6}$

Step-by-step explanation:

$\int\limits^4_3 {x^2-x} \, dx$

= [ $\frac{x^3}{3}$ - $\frac{x^2}{2}$ ] ← evaluate for upper limit - lower limit

= ( $\frac{64}{3}$ - $\frac{16}{2}$ ) - ( $\frac{27}{3}$ - $\frac{9}{2}$ )

= $\frac{64}{3}$ - 8 - 9 + $\frac{9}{2}$

= $\frac{155}{6}$ - 17

= $\frac{53}{6}$

5 0

3 years ago

Last week, the members of a weight loss support group collectively weighed 1,580 pounds. They just weighed in again and are now

rosijanka [135]

Answer: 1,501

Step-by-step explanation:

100% = 1,580

100% - 5% lost

95%

1580 x .95 = 1501

3 0

3 years ago

A researcher wishes to estimate the number of households with two cars. How large a sample is needed in order to be 99% confiden

solmaris [256]

Answer:

Sample size n = 1382

so correct option is D) 1382

Step-by-step explanation:

given data

confidence level = 99 %

margin of error = 3%

probability = 25 %

to find out

How large a sample size needed

solution

we know here P = 25 %

so 1 - P = 1 - 0.25

1 - P = 0.75

and we know E margin of error is 0.03 so value of Z for 99%

α = 1 - 99% = 1 - 0.99

α = 0.01

and $\frac{\alpha}{2}$ = $\frac{0.01}{2}$

$\frac{\alpha}{2}$ = 0.005

so Z is here

$Z_(\frac{\alpha}{2})$ = 2.576

sample size will be

Sample size n = $(\frac{(Z_(\frac{\alpha}{2})}{E})^2 * P * (1-P)$

put here value

Sample size n = (\frac{2.576}{0.03})^2 * 0.25 * 0.75

Sample size n = 1382

so correct option is D) 1382

3 0

3 years ago