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vitfil [10]
2 years ago
9

Brian buys 5 cookies and 1 bottle of water. The total cost is $8.75. If the bottle of water costs $1.25, which equation can be u

sed to determine the cost of 1 cookie?
A, 5c+1.25=8.75
B, 5(1.25)+c
C, 5c+8.75=1.25
D, c+1.25=8.75
Mathematics
2 answers:
lord [1]2 years ago
7 0

Answer:

A. 5c + 1.25 = 8.75

Step-by-step explanation:

Where c is cost of one cookie

5c + 1.25 = 8.75

Subtract 1.25 from both sides

5c + 1.25 - 1.25 = 8.75 - 1.25

5c = 7.5

Divide both sides by 5

c = 7.5/5

c = 1.5

= $1.5

One cookie cost $1.5

yawa3891 [41]2 years ago
6 0

Answer:

A

(i may be wrong)

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At a rally attended by 520 people, 30% were fantes, 25% were ewes, 15% were nzemas, 20% were gas and the rest were gonjas. How m
mariarad [96]

Answer:

No. of gonjas = 52

No. of more nzemas than fantes = 78

Step-by-step explanation:

Total no. of people = 520

No. of fantes = \frac{30}{100} × 520

No. of fantes = 156

No. of ewes = \frac{25}{100} × 520 = 130

No. of nzemas = \frac{15}{100} × 520 = 78

No. of gas = \frac{20}{100} × 520 = 104

No. of gonjas = 520 - (156 + 130 + 78 + 104) = 52

No. of fantes = 156

No. of nzemas = 78

No. of more nzemas than fantes = 156 - 78 = 78

Pie chart of the following problem is shown below.

5 0
2 years ago
Read 2 more answers
A home’s value increases at an average rate of 5.5% each year. The current value is $120,000. What function can be used to find
Wewaii [24]

Answer:

y=120,000(1.055)^x

Step-by-step explanation:

we know that

The equation of a exponential growth function is given by

y=a(1+r)^x

where

y is the value of the home

x is the number of years

a is the initial value

r is the rate of change

we have

a=\$120,000\\r=5.5\%=5.5/100=0.055

substitute

y=120,000(1+0.055)^x

y=120,000(1.055)^x

7 0
2 years ago
44 students completed some homework and the histogram shows information about the times taken. Work out an estimate of the inter
Vilka [71]

Answer:

q1=11

q3= 33

Step-by-step explanation:

The data set has 44 number of students. The first quartile is 25 %  of the numbers in the data set . So

25 % of 44 = 25/100 * 44= 0.25 *44 = 11

So the first quartile lies at 11.

Similarly the third quartile lies at the 75 % of the numbers of the data set . So

75 % of 44 = 75/100 * 44= 0.75 *44 = 33

So the third quartile lies at 33.

5 0
2 years ago
Ethan sells e candy bars for $2.50 apiece and Chloe sells c candy bars for $2.00 apiece to raise
natta225 [31]
2.50x = Ethan’s total sold -6 (15/2.50=6)
2x = Chloe’s total sold
2.50(12) = 30 - 6 = 24
2(12) = 24
They both sold 12 candy bars
8 0
2 years ago
Read 2 more answers
Tesla wanted to determine the average miles per kWh that their vehicles get across all models and variations. They took a sample
LiRa [457]

Answer:

\mu_{\bar x} = \mu = E(X) =30KWh

\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{100}}=0.3

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

For this case we select a sample of n =100

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

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

So then the sample mean would be:

\mu_{\bar x} = \mu = E(X) =30KWh

And the standard deviation would be:

\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{100}}=0.3

6 0
2 years ago
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