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3 years ago

A recipe calls for 2 1/2 cups of flour to make 24 cookies how many cups of flour would be required to bake 15 dozen cookies

Mathematics

2 answers:

Citrus2011 [14]3 years ago

7 0

15 dozen = 180 cookies

180 ÷ 24 = 7.5

7.5 x 2.5 = 18 3/4 cups of flour

trasher [3.6K]3 years ago

5 0

let 'X' represent the cups of flour required to make 15 dozen cookies.

15: X=2 : (5/2) (since 2 1/2=5/2)

x=75/4

×=18 3/4 cups

18 3/4 cups are required to bake 15 dozen of cookies.

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Evaluate each expression below. Click on "Undefined" as needed. 0/9= 3÷0 =

damaskus [11]

0/9 = 0 and 3/0 = undefined. Any number divided by 0 is undefined, while 0 divided by any number is just 0.

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3 years ago

The function f(x) varies inversely with x and f(x)= -10 when x = 5

wlad13 [49]

F(x)= -10
f(5)=-10
f=-10 divided by 5
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3 years ago

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

Ivanshal [37]

Answer:

a) $\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

b) $z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

8 0

3 years ago

Lizzie rolls two dice. What is the probability that the sum of the dice is:

zhenek [66]

Answer:

$A.\ \dfrac{1}{3}\\B.\ \dfrac{5}{12}\\C.\ \dfrac{7}{36}\\$

Step-by-step explanation:

Total outcomes possible: 36

A. Divisible by 3

Possible options are:

3, 6, 9 and 12.

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Possible outcomes for 9 are: {(3,6), (4,5), (5,4),(6,3)} Count 4

Possible outcomes for 12 are: {(6,6)} Count 1

Total count = 2 + 5 + 4 + 1 = 12

Probability of an event E can be formulated as:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

$P(A) = \dfrac{12}{36} = \dfrac{1}{3}$

B. Less than 7:

Possible sum can be 2, 3, 4, 5, 6

Possible cases for sum 2: {(1,1)} Count 1

Possible cases for sum 3: {(1,2), (2,1)} Count 2

Possible cases for sum 4: {(1,3), (3,1), (2,2)} Count 3

Possible cases for sum 5: {(1,4), (2,3), (3,2),(4,1)} Count 4

Possible cases for sum 6: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Total count = 1 + 2 + 3 + 4 + 5 = 15

$P(B) = \dfrac{15}{36} = \dfrac{5}{12}$

C. Divisible by 3 and less than 7:

$P(A \cap B) = \dfrac{n(A\cap B)}{\text{Total Possible outcomes}}$

Here, common cases are:

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

$P(A \cap B) = \dfrac{7}{\text{36}}$

5 0

2 years ago

Which of the following is an informal proof of the given conjecture? Conjecture: All squares are similar

tester [92]

The 3rd one because of the laws and newton’s ways

7 0

3 years ago