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Colt1911 [192]
2 years ago
10

lee is making chocolate chip cookies and the recipe calls for 3/4 cup of chocolate chips. if she wants to make 2/3 of the recipe

, what fraction of a chip of chocolate chips will she need? explain.
Mathematics
2 answers:
Marysya12 [62]2 years ago
7 0

<span>
First, (again lol) you multiply the amount of chocolate chips the recipe calls for by the amount Lee wants to make.</span>

<span />so its 3/4 * 2/3

<span> =</span>

<span>6/12</span>

<span>1/2</span>

<span>Lee will need a fraction of 1/2 of a chip of chocolate chips for the recipe</span>

Vesnalui [34]2 years ago
4 0

Answer:

\frac{1}{2} cups

Step-by-step explanation:

Lee is making chocolate chip cookies and recipe requires \frac{3}{4} cups of chips

Then for \frac{2}{3} fraction of the recipe will require chocolate chips = \frac{3}{4} × \frac{2}{3}

        = \frac{1}{2} cups

Therefore, Lee will require  \frac{1}{2} cups of chocolate chips.

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Triangle A B C is shown. Angle B C A is a right angle. The length of hypotenuse A B is 5, the length of B C is 3, and the length
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The side opposite to angle B is the side that does not contact with angle B.

In this attached image, you can see better that sides AB and BC is in contact with angle B. So, the opposite side to angle B is AC.

Therefore, the lenght of the side opposite to angle B is 4 units.

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3,298,076 in expanded form
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3,000,000+200,000+90,000+8,000+70+6
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1. Kellie and her sister Ashley are training for a marathon. Kellie ran 10 mi in 75 min. Ashley ran 15 mi in 120 min. Which stat
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(a) Kellie’s minutes-per-mile pace was faster than Ashley’s minutes-per-mile pace.

True

10/75 = 0.14

15/120 = 0.13

(b) Kellie ran 8 mph.

True

10 ÷ 1.15 = 8

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False

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8 0
1 year ago
Read 2 more answers
Three assembly lines are used to produce a certain component for an airliner. To examine the production rate, a random sample of
nikitadnepr [17]

Answer:

a) Reject H₀

b) [0.31; 3.35]

Step-by-step explanation:

Hello!

a) The objective of this example is to compare if the population means of the production rate of the assembly lines A, B and C. To do so the data of the production of each line were recorded and an ANOVA was run using it.

The study variable is:

Y: Production rate of an assembly line.

Assuming that the study variable has a normal distribution for each population, the observations are independent and the population variances are equal, you can apply a parametric ANOVA with the hypothesis:

H₀ μ₁= μ₂= μ₃

H₁: At least one of the population means is different from the others

Where:

Population 1: line A

Population 2: line B

Population 3: line C

α: 0.01

This test is always one-tailed to the right. The statistic is the Snedecor's F, constructed as the MSTr divided by the MSEr if the value of the statistic is big, this means that there is a greater variance due to the treatments than to the error, this means that the population means are different. If the value of F is small, it means that the differences between populations are not significant ( may differ due to error and not treatment).

The critical region is:

F_{k-1;n-k; 1-\alpha } = F_{2;15; 0.99} = 6.36

If F ≥ 3.36, the decision is to reject the null hypothesis.

Looking at the given data:

F= \frac{MSTr}{MSEr}= 11.32653

With this value the decision is to reject the null hypothesis.

Using the p-value method:

p-value: 0.001005

α: 0.01

The p-value is less than the significance level, the decision is to reject the null hypothesis.

At a level of 5%, there is significant evidence to say that at least one of the population means of the production ratio of the assembly lines A, B and C is different than the others.

b) In this item, you have to stop paying attention to the production ratio of the assembly line A to compare the population means of the production ratio of lines B and C.

(I'll use the same subscripts to be congruent with part a.)

The parameter to estimate is μ₂ - μ₃

The populations are the same as before, so you can still assume that the study variables have a normal distribution and their population variances are unknown but equal. The statistic to use under these conditions, since the sample sizes are 6 for both assembly lines, is a pooled-t for two independent variables with unknown but equal population variances.

t=  (X[bar]₂ - X[bar]₃) - ( μ₂ - μ₃) ~t_{n_2+n_3-2}

Sa√(1/n₂+1/n₃)

The formula for the interval is:

(X[bar]₂ - X[bar]₃) ± t_{n_2+n_3-2; 1 - \alpha /2}* Sa\sqrt{*\frac{1}{n_2} + \frac{1}{n_3} }

Sa^{2} = \frac{(n_2-1)*S_2^2+ (n_3-1)*S_3^2}{n_2+n_3-2}

Sa^{2} = \frac{(5*0.67)+ (5*0.7)}{6+6-2}

Sa^{2} = 0.685

Sa= 0.827 ≅ 0.83

t_{n_2+n_3-2;1-\alpha /2}= t_{10;0.995} = 3.169

X[bar]₂ = 43.33

X[bar]₃ = 41.5

(43.33-41.5) ± 3.169 * *0.83\sqrt{*\frac{1}{6} + \frac{1}{6} }

1.83 ± 3.169 * 0.479

[0.31; 3.35]

With a confidence level of 99% you'd expect that the difference of the population means of the production rate of the assemly lines B and C.

I hope it helps!

8 0
1 year ago
(b) A department store has 7,000 charge accounts. The comptroller takes a random sample of 36 of
galben [10]

Answer:

a.0.8664

b. 0.23753

c. 0.15866

Step-by-step explanation:

The comptroller takes a random sample of 36 of the account balances and calculates the standard deviation to be N42.00. If the actual mean (1) of the account balances is N175.00, what is the probability that the sample mean would be between

a. N164.50 and N185.50?

b. greater than N180.00?

c. less than N168.00?

We solve the above question using z score formula

z = (x-μ)/σ/√n where

x is the raw score,

μ is the population mean = N175

σ is the population standard deviation = N42

n is random number of sample = 36

a. Between N164.50 and N185.50?

For x = N 164.50

z = 164.50 - 175/42 /√36

z = -1.5

Probability value from Z-Table:

P(x = 164.50) = 0.066807

For x = N185.50

z = 185.50 - 175/42 /√36

z =1.5

Probability value from Z-Table:

P(x=185.50) = 0.93319

Hence:

P(x = 185.50) - P(x =164.50)

= 0.93319 - 0.066807

= 0.866383

Approximately = 0.8664

b. greater than N180.00?

x > N 180

Hence:

z = 180 - 175/42 /√36

z = 5/42/6

z = 5/7

= 0.71429

Probability value from Z-Table:

P(x<180) = 0.76247

P(x>180) = 1 - P(x<180) = 0.23753

c. less than N168.00?

x < N168.

z = 168 - 175/42 /√36

z = -7/42/6

z = -7/7

z = -1

Probability value from Z-Table:

P(x<168) = 0.15866

4 0
1 year ago
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