All categories

2 years ago

7. Marie can make 12 small mango pies for every 10 mangoes. How many pie

Mathematics

1 answer:

Marysya12 [62]2 years ago

8 0

Answer:

D. Marie can make 60 pies.

Step-by-step explanation:

In this problem, we know that amount of small mango pies is directly proportional to the amount of mangoes. The amount needed for preparing an amount of pies is calculated by simple rule of three:

A. Marie can make 30 pies.

$x = \frac{10\,mangoes}{12\,pies}\times 30\,pies$

$x = 25\,mangoes$

25 mangoes are required for producing 30 pies.

B. Marie can make 40 pies.

$x = \frac{10\,mangoes}{12\,pies}\times 40\,pies$

$x = 33.333\,mangoes$

34 mangoes are required for producing 40 pies.

C. Marie can make 50 pies.

$x = \frac{10\,mangoes}{12\,pies}\times 50\,pies$

$x = 41.667\,mangoes$

42 mangoes are required for producing 50 pies.

D. Marie can make 60 pies.

$x= \frac{10\,mangoes}{12\,pies}\times 60\,pies$

$x = 50\,mangoes$

50 mangoes are required for producing 60 pies.

The complete statement is: Marie can make 12 small mango pies for every 10 mangoes.How many pies can she make with 50 mangoes.

Hence, correct answer is D.

You might be interested in

Which expression is equivalent to 2+3

disa [49]

Answer:

Step-by-step explanation:

1. 9x^5

2. 9x^6

3. 27x^5

4. 27x^6

3 0

3 years ago

Read 2 more answers

Find the greatest common factor (GCF) of the set of numbers. 21, 30, 44

Vitek1552 [10]

Step-by-step explanation:

Answer attached :)

Hope it helps :D

7 0

3 years ago

Joe has 5 more than twice as many calculators as John. Write an algebraic expression. *

yarga [219]

Answer:

B. 5 + 2p

Step-by-step explanation:

Let's simplify the problem:

Joe has 5 + more than 2(P) as many calculators as John.

*We'll be using Addition and Multiplication operations.

If Joe has 5 more than 2 times the amount of calculators, the algebraic expression will be:

$5 + 2p$

p represents John's amount of calculators.

5 + (5 more than) 2(P) Twice the amount of calculators John has.

7 0

2 years ago

Twenty students from Sherman High School were accepted at Wallaby University. Of

nydimaria [60]

Answer:

Data provide convincing evidence of a difference in SAT scores between students with and without a military scholarship is explained below in details.

Step-by-step explanation:

This is a quiz of 2 autonomous groups. The population model differences are not understood. it is a two-tailed examination. Let w be the index for scores of students with army research and o be the index for scores of students without army research.

Therefore, the population means would be μw and μo.

The irregular variable is x w - xo = variation in the sample mean records of students with military accomplishments and students without.

For students with military accomplishments,

n = 8

Mean = (850 + 925 + 980 + 1080 + 1200 + 1220 + 1240 + 1300)/8

Mean = 1099.375

Standard deviation = √(summation(x - mean)/n

Summation(x - mean) = (850 - 1099.375)^2 + (925 - 1099.375)^2 + (980 - 1099.375)^2 + (1080 - 1099.375)^2 + (1200 - 1099.375)^2 + (1220 - 1099.375)^2 + (1240 - 1099.375)^2 + (1300 -1099.375)^2 = 191921.875

Standard deviation = √(191921.875/8 = 154.89

For students without military scholarship,

n = 12

Mean = (820 + 850 + 980 + 1010 + 1020 + 1080 + 1100 + 1120 + 1120 + 1200 + 1220 + 1330)/12

Mean = 1073.83

Summation(x - mean) = (820 - 1073.83)^2 + (850 - 1073.83)^2 + (980 - 1073.83)^2 + (1010 - 1073.83)^2 + (1020 - 1073.83)^2 + (1080 - 1073.83)^2 + (1100 - 1073.83)^2 + (1120 - 1073.83)^2 + (1120 - 1073.83)^2 + (1200 - 1073.83)^2 + (1220 - 1073.83)^2 + (1330 - 1073.83)^2 = 238199.4268

Standard deviation = √(238199.4268/12 = 140.89

We would set up the hypothesis.

The null hypothesis is

H0 : μw = μo H0 : μw - μo = 0

The alternative hypothesis is

Ha : μw ≠ μo Ha : μw - μo ≠ 0

Since sample standard deviation is recognized, we would analyis the examination statistic by using the t examination. The formula is

(xw - xo)/√(sw²/nw + so²/no)

From the information given,

xw = 1099.375

xo = 1073.83

sw = 154.89

so = 140.89

nw = 8

no = 12

t = (1099.375 - 1073.83)/√(154.89²/8 + 140.89²/12)

t = 0.37

The formula for determining the degree of freedom is

df = [sw²/nw + so²/no]²/(1/nw - 1)(sw²/nw)² + (1/no - 1)(so²/no)²

df = [154.89²/8 + 140.89²/12]²/(1/8 - 1)(154.89²/8)² + (1/12 - 1)(140.89²/12)² = 21650688.37/1533492.15

df = 14

We would get the probability count from the t test calculator. It becomes

p value = 0.72

Since the level of importance of 0.05 < the p value of 0.72, we would not neglect the null hypothesis.

Therefore, these data do not present an acceptable indication of a difference in SAT scores between students with and without a military scholarship.

Part B

The formula for getting the confidence interval for the difference of two population means is expressed as

z = (xw - xo) ± z ×√(sw²/nw + so²/no)

For a 95% confidence interval, the z score is 1.96

xw - xo = 1099.375 - 1073.83 = 25.55

z√(sw²/nw + so²/no) = 1.96 × √(154.89²/8 + 140.89²/12) = 1.96 × √2998.86 + 1654.17)

= 133.7

The confidence interval is

25.55 ± 133.7

6 0

3 years ago

a tune up automotive facility marks up its parts by 45%. suppose that the facility charges its customers 2.32 for each spark plu

V125BC [204]

Answer:

$1.6

Step-by-step explanation:

The price of $2.32 already includes the 45% mark up hence taking the original cost to be x, it means the cost of $2.32 is 145% of the original cost

Using the concept of cross multiplication we say that if

1.45 is equivalent to $2.32

1 is equivalent to x

$x=\frac {2.32}{1.45}=1.6$

Therefore, the company incurs $1.6 for each plug

4 0

3 years ago