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faust18 [17]
2 years ago
14

A box of cereal states that there are 75 calories in a three fourths 3/4​-cup serving. What is the unit rate for calories per​ c

up? How many calories are there in 2 cups of the​ cereal?
Mathematics
1 answer:
Dimas [21]2 years ago
7 0

Answer:

a) The unit rate for calories per​ cup is 100 calories per cup

b) 200 calories in 2 cups of cereal.

Step-by-step explanation:

A box of cereal states that there are 75 calories in a three fourths 3/4​-cup serving.

a) What is the unit rate for calories per​ cup?

This is calculated as:

3/4 cup = 75 calories

1 cup = x

Cross Multiply

3/4x = 75 calories × 1

x = 75 calories /3/4 cup

x = 75 ÷ 3/4

x = 75 × 4/3

x = 100 calories per cup

The unit rate for calories per​ cup is 100 calories per cup

b) How many calories are there in 2 cups of the​ cereal?

From the above question

3/4 cup = 75 calories

2 cups = x

Cross Multiply

3/4x = 75× 2

x = 150 ÷ 3/4

x = 150 × 4/3

x = 200 calories

There are 200 calories in 2 cups of cereal.

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When data are collected using a quantitative, ratio variable, what is true about a frequency distribution that summarizes the da
bazaltina [42]

When data are collected using a quantitative, ratio variable, a frequency distribution that summarizes the data is the upper and lower class limits must be calculated is the correct answer.

In this question,

A frequency distribution is used to display the number of observations within a particular interval. It groups data into categories, and records the number of observations in each category.

Frequency distributions are visual displays that organize and present frequency counts so that the information can be interpreted more easily. It is a graphical or tabular representation of data that shows the number of observations within a given value.

It provides the information of the number of occurrences (frequency) of distinct values distributed within a given period of time or interval, in a list, table, or graphical representation.

From the definition, when data are collected using a quantitative, ratio variable, a frequency distribution that summarizes the data is the upper and lower class limits must be calculated is the correct answer.

Learn more about frequency distribution here

brainly.com/question/14866043

#SPJ4

The complete question is as follows,

When data are collected using a quantitative, ratio variable, what is true about a frequency distribution that summarizes the data?

A. A pie chart can be used to summarize the data, B. Upper and lower class limits must be calculated C. Number of classes is equal to the number of variable's values. D. The "5 to the k rule" can be applied.

6 0
1 year ago
Can someone help :0 not sure how to solve but I'm doing algebra if that helps
Rus_ich [418]

Answer:<em>r = d/t</em>

Step-by-step explanation:

1. If we are solving for <em>r, </em>we need to have <em>r</em> on one side of the equation.

2. Divide by t on both sides = d/t=r

8 0
3 years ago
Two surfers and statistics students collected data on the number of days on which surfers surfed in the last month for 30 longbo
Alisiya [41]

Answer:

Do not reject H0. The mean days surfed for longboarders is significantly larger than the mean days surfed for all shortboarders

Step-by-step explanation:

The null hypothesis is that  the mean days surfed for all long boarders is larger than the mean days surfed for all short boarders

H0:  μL > μs      against the claim Ha:  μL≤ μs

the alternate hypothesis is the mean days surfed for all long boarders isless or equal to  the mean days surfed for all short boarders (because long boards can go out in many different surfing conditions)

The test statistic is

t= x1- x2/  √s1/n1+ s2/n2

1) Calculations

Longboards

Mean

ˉx=∑x/n=4+8+9+4+9+7+9+6+6+11+15+13+16+12+10+12+18+20+15+10+15+19+21+9+22+19+23+13+12+10/30

=377/30

=12.5667

Longboard Variance S2=[∑dx²-(∑dx)²/n]/n-1

=[831-(-13)²/30]/29

=831-5.6333/29

=825.3667/29

=28.4609

Shortboard Mean

ˉx=∑x/n=6+4+6+6+7+7+7+10+4+6+7+5+8+9+4+15+13+9+12+11+12+13+9+11+13+15+9+19+20+11/30

=288/30

=9.6

Shortboard Variance S2=[∑x²-(∑x)²/n]/n-1

=[ 3270-(288)2/30]/29

=3270-2764.8/29

=505.2/29

=17.4207

2) Putting values in the test statistic

t=|x1-x2|/√S²1/n1+S²2/n2

t =|12.5667-9.6|/√28.4609/30+17.4207/30

t =|2.9667|/√0.9487+0.5807

t=|2.9667|/√1.5294

t=|2.9667|/1.2367

t=2.3989

3) Degree of freedom =n1+n2-2=30+30-2=58

4) The critical region is t ≤ t(0.05) (58) =1.6716

5) Since the calculated t= 2.4 does not fall in the critical region t(0.05) (58)  ≤ 1.6716 we do not reject H0.

The p-value is 0.008969. The result is significant at p <0 .05.

6 0
3 years ago
A spinner has equal regions numbered 1 through 20. What is the probability that the spinner will stop on an odd number or a mult
klio [65]

Answer:

<em>Probability that the spinner will stop on an odd number or a multiple of 5 is </em><em>0.6</em>

Step-by-step explanation:

Probability = \frac{Required outcomes}{Total possible outcomes}

We are given the equal regions numbered from 1 through 20 which means that our total possible outcomes are 20

<em>Total possible outcomes: 20</em>


<em>Outcomes that spinner will stop on an odd number, n(Odd): 10</em>

1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Probability of spinner stoping on Odd number:

P(Odd) = \frac{n(Odd)}{Total} = \frac{10}{20} = \frac{1}{2} = 0.5


Outcomes that spinner will stop on a multiple of 5, n(5): 4

5, 10, 15, 20

Probability of spinner stoping on multiple of 5:

P(5) = \frac{n(5)}{Total} = \frac{4}{20} = \frac{1}{5} = 0.2

Odd numbers which are a multiple of 5 are: 5 and 15

So,

P(Odd and 5) = \frac{2}{20}=\frac{1}{10}=0.1

Thus Probability of spinner stopping at odd number or a multiple of 5 becomes:

P(Odd or 5) = P(Odd) + P(5) - P(Odd and 5) = 0.5 + 0.2 - 0.1 = 0.6

7 0
3 years ago
Find the slope and y-intercept of the line <br> 6x + 2y = –88
ira [324]
Y=mx+b
6x+2y=-88
2y=-88-6x
y=-44-3x
y=-3x-44
m is the slope
b is the y-intercept
4 0
3 years ago
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