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Romashka [77]
1 year ago
12

Mandy is making stationery sets from 16 sheets of paper and 12 envelopes. If she wants all the sets to be identical without any

paper or envelopes left over, what is the greatest number of sets Mandy can make
Mathematics
1 answer:
LekaFEV [45]1 year ago
6 0

Answer:

  4 sets

Step-by-step explanation:

The greatest common factor (GCF) of 16 and 12 is 4. This is the largest number that evenly divides 12 and 16. It can be found by looking at the factors of those numbers:

  12 = 4·3

  16 = 4·4

Or it can be found by seeing if the difference of the numbers (4) evenly divides them both. It does, so that is the GCF.

The greatest number of sets Mandy can make is 4.

__

Each of Mandy's 4 sets will have 4 sheets of paper and 3 envelopes.

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Trava [24]

Answer:

C) Both statements could be correct. RST could be the result of two translations of ABC. TSR could be the result of a reflection and a translation of ABC.

Step-by-step explanation:

When naming congruent shapes, the <u>orders of the congruent vertex letters need to be the same</u>.

Since these are isosceles triangles, the base angles are the same:

m∠R = m∠T = m∠A = m∠C

Therefore the congruency statement can be written two different ways.

ΔABC ≅ ΔRST

ΔABC ≅ ΔTSR

Both statements could be correct.

Choosing between B) and C):

To move ΔABC to where ΔRST or ΔTSR is, you could either:

i) Translate 6 units to the left, and translate 3 units down

ii) Reflect across the y-axis, and translate 3 units down

It can be the result of two translations or a reflection and a translation.

In the result, the base side RT is on the bottom of the shape, like side AC. If you rotated the shape, the base side would not be on the bottom. Therefore B) is incorrect.

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2 years ago
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Adamo saved 13 pictures oh his digital camera for a total of 12021.1kb. He deleted 32 pictures for a total of 29590.4kb If there
rosijanka [135]

First solve the original number of pictures. The original number of picture is 108 + 32 – 13 which is equal to 127 pictures. Then solve the size of each picture,

12021.1 kb / 13 pic = 924.7 kb / pic

So the original storage is 127 pic ( 924.7 kb / pic) = 117,436.9 kb

5 0
2 years ago
Consumer Reports scored cars on quality and cost. Spearman's rank correlation coefficient was calculated using Statdisk and the
melomori [17]

Answer:

d.There is insufficient evidence to conclude that the quality and price of a car are associated. There were ten cars used in the sample.

Step-by-step explanation:

Hello!

You have two variables X₁: quality score of a car and X₂: the price of a car.

It was analyzed id there is an association between the quality and the price.

The null hypothesis of a Spearman's rank correlation test is:

H₀: There is no association between the quality and the price of cars.

The researcher failed to reject the null hypothesis which means that there is no association between the variables of interest.

The sample size is listed in the output n= 10 consumer reports.

I hope you have a SUPER day!

8 0
1 year ago
Jayla buys and sells vintage clothing. She bought two blouses for $25.00 each and later sold them for $38.00 each. She bought th
4vir4ik [10]
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2 years ago
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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
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