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2 years ago

Steve got 57 questions correct on his exam. This was 95%. How many questions were there on the exam?

2 answers:

8 0

The answer is D, because if you use common sense you would see that it wouldn’t take over 100 more questions to give Steve 100%

lilavasa [31]2 years ago

3 0

Answer: 60 questions, option D is correct

Step-by-step explanation:

Divide the number of correct questions by the percentage

=57/95% ; 95% = 0.95

=57/0.95

= 60

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Emily and a friend bought two tickets to see a soccer game. Each ticket cost $8.25. The friends paid a total of $24.50, which in

solniwko [45]

$2\cdot8.25=16.5\\\\24.5-16.5=8\\\\8:2=4\\\\\frac{4}{8.25}\cdot100\%=\\\\\frac{400}{8.25}\%=\\\\48.48\%$

6 0

2 years ago

An octopus weighted 5/6 kilogram. After two weeks, it's weight has increased by 3/10 kilogram. But afterwards, it lost 1/5 kilog

Alex777 [14]

Answer

14/15 kg

Step-by-step explanation:

Rewriting our equation with parts separated

=1+215−15

Solving the fraction parts

2/15−15=?

Find the LCD of 2/15 and 1/5 and rewrite to solve with the equivalent fractions.

LCD = 15

2/15−3/15=−1/15

Combining the whole and fraction parts

1−1/15=14/15

Solution by Formulas

converting mixed numbers to fractions, our initial equation becomes,

17/15−15

Applying the fractions formula for subtraction,

=(17×5)−(1×15)15×5

=85−15/75

=70/75

Simplifying 70/75, the answer is

=14/15

5 0

2 years ago

The average annual cost of the first year of owning and caring for a large dog is $1,843 (US News and World Report, September 9,

ICE Princess25 [194]

Answer:

a. Error is 0.14

b. between 1913.68 to 1772.32

Step-by-step explanation:

Requirement a)

The margin of error is a statistic expressing the amount of random sampling error in a survey's results.

Given,

Size of the sample n = 50

We know, At 95% confidence interval,

The margin of error,

= 0.98/√n

= 0.98/√50

= 0.98/7.07

= 0.13859 ~ 0.14

So the margin of error at 95% confidence level is 0.14.

Requirement b)

Given,

Standard deviation σ = 255

Population mean z = 1814

Size of the sample n = 50

We know, At 95% confidence interval,

(z-µ)/(σ/√n) < Z_0.05

= - Z_0.05 < (z-µ)/(σ/√n) < Z_0.05

= - 1.96 * σ/√n < (z-µ)/(σ/√n)* σ/√n < 1.96 * σ/√n [ in two tail Z test the value of Z_0.05 is 1.96 ]

= - 1.96 * 255/√50 < Z-µ < 1.96 * 255/√50

= -70.6828 < Z-µ < 70.6828

= 70.6828 > µ - Z > -70.6828

= 70.6828 + Z > µ > -70.6828 + Z

= 70.6828 + 1814 > µ > -70.6828 + 1814

=1913.68 > µ > 1772.32

So, the sample mean would be between 1913.68 to 1772.32.

7 0

2 years ago

The amount of rhubarb in the original recipe is 3 1/2 cups. Using what you know of whole numbers and what you know of fractions,

Alexxx [7]

The easiest way, I think, is to convert the mixed number into an improper fraction, then multiply by 3.
3 1/2 = 7/2
7/2 · 3 = 21/2
now just change the improper fraction back to a mixed number by dividing and putting the remainder into fraction form
21/2 = 10 1/2

You could also multiply the whole number by 3 and the fraction by 3, ending up with 9 3/2, but then have to convert the improper fraction into a mixed number
3/2 = 1 1/2
then add the numbers together
9 + 1 1/2 = 10 1/2
either way works, whatever is easiest for you.

8 0

2 years ago

a photo collage consists of seven equal sized square photo arranged in a row to form a rectangle if the total area of the colleg

Keith_Richards [23]

Answer:

Step-by-step explanation:

I have attached a written explanation to help :).

Let's say each side of the square = $x$