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Veronika [31]
3 years ago
5

0.9 of Karen's Science score is equal to 0.75 of her Math score and 0.8 of her Chemistry score. Given that Karen scored 96 marks

for her Math, what was her total score of the 3 subjects?
Mathematics
1 answer:
DochEvi [55]3 years ago
5 0

You have to formulate equations for this problem.

Let S = Science score

      M = Math score

       C = Chemistry score

To illustrate the given:

0.9S = 0.75M

0.9S = 0.8C

You are given that Karen’s Math score is 96 marks. You have to substitute the Math score to the first equation.

0.9S = 0.75(96)

0.9S = 72

S = 80

Therefore, Karen’s Science score is 80. Now, you have to substitute the Science score to the second equation.

0.9(80) = 0.8C

0.8C = 72

C = 90

So, Karen’s Chemistry score is 90.

Therefore, the total score of the 3 subjects is 266 (96 + 80 + 90).

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masha68 [24]

Answer:

32

Step-by-step explanation:

6 0
3 years ago
Yesterday Mike bought 2 gallons of regular gasoline and 3 gallons of premium gasoline at a gas station for $13.60 today he bough
larisa [96]

Answer:

Cost of 1 gallon of Regular gasoline is $2.45 and  Cost of 1 gallon of Premium gasoline is $2.90.

Step-by-step explanation:

Let the Cost of Regular gasoline be 'x'.

Let the Cost of Premium gasoline be 'y'.

Given:

Amount of regular gasoline bought yesterday = 2 gallons

Amount of premium gasoline bought yesterday = 3 gallons

Total Cost of yesterday = $13.60

Now we know that Total Cost of yesterday is equal to sum of Amount of regular gasoline bought yesterday multiplied by Cost of Regular gasoline and Amount of Premium gasoline bought yesterday multiplied by Cost of premium gasoline.

Framing in equation form we get;

2x+3y=13.60 \ \ \ \ equation\ 1

Also Given:

Amount of regular gasoline bought today = 3 gallons

Amount of premium gasoline bought Today = 4 gallons

Total Cost of Today = $18.95

Now we know that Total Cost of Today is equal to sum of Amount of regular gasoline bought Today multiplied by Cost of Regular gasoline and Amount of Premium gasoline bought Today multiplied by Cost of premium gasoline.

Framing in equation form we get;

3x+4y=18.95 \ \ \ \ equation\ 2

Now Multiplying equation 1 by 3 we get;

2x+3y=13.60\\\\3(2x+3y)=13.60\times3\\\\6x+9y= 40.80 \ \ \ \ \ equation\ 3

Now Multiplying equation 2 by 2 we get;

3x+4y=18.95\\\\2(3x+4y)=18.95\times2\\\\6x+8y= 37.90 \ \ \ \ \ \ equation\ 4

Subtracting equation 4 from equation 3 we get;

(6x+9y)- (6x+8y)= 40.80-37.90\\\\6x+9y-6x-8y= 2.9\\\\y=\$2.90

Substituting the value of y in equation 1 we get;

2x+3y=13.60\\\\2x+3\times2.90 =13.60\\\\2x+8.7=13.6\\\\2x=13.6-8.7\\\\2x=4.9\\\\x=\frac{4.9}{2} =\$2.45

Hence Cost of 1 gallon of Regular gasoline is $2.45 and  Cost of 1 gallon of Premium gasoline is $2.90.

4 0
3 years ago
Please help me ASAP!!! Thank you!!
Airida [17]

Answer:

Y≤-3/4x+2

Step-by-step explanation:

less than because the bottom of the line is shaded

equal to because the line is solid

slope you go down three and go right four which is negative

y intercept is where x=0

5 0
3 years ago
The level of nitrogen oxides (NOX) in a exhaust of cars of a particular model varies normally with mean 0.25 grams per miles and
antoniya [11.8K]

Answer:

a) 15.87% probability that a single car of this model fails to meet the NOX requirement.

b) 2.28% probability that the average NOX level of these cars are above 0.3 g/mi limit

Step-by-step explanation:

We use the normal probability distribution and the central limit theorem to solve this question.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 0.25, \sigma = 0.05

a. What is the probability that a single car of this model fails to meet the NOX requirement?

Emissions higher than 0.3, which is 1 subtracted by the pvalue of Z when X = 0.3. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{0.3 - 0.25}{0.05}

Z = 1

Z = 1 has a pvalue of 0.8417.

1 - 0.8413 = 0.1587.

15.87% probability that a single car of this model fails to meet the NOX requirement.

b. A company has 4 cars of this model in its fleet. What is the probability that the average NOX level of these cars are above 0.3 g/mi limit?

Now we have n = 4, s = \frac{0.05}{\sqrt{4}} = 0.025

The probability is 1 subtracted by the pvalue of Z when X = 0.3. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{0.3 - 0.25}{0.025}

Z = 2

Z = 2 has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% probability that the average NOX level of these cars are above 0.3 g/mi limit

4 0
3 years ago
ABC is a right triangle. If AB = 3 and AC = 7, find BC. Leave your answer in simplest radical form.
aivan3 [116]
1. Suppose that BC is hypotenuse, then by Pythagorean theorem:
BC^2=AC^2+AB^2, \\ BC^2=3^2+7^2, \\ BC^2=9+49, \\ BC^2=58, \\ BC= \sqrt{58}.
2. If BC is not hypotenuse, then BC is a leg and AC is hypotenuse (because AC>AB). Hence by Pythagorean theorem:
AC^2=AB^2+BC^2, \\ 7^2=3^2+BC^2, \\ BC^2=49-9, \\ BC^2=40, \\ BC= \sqrt{40} =2 \sqrt{x10}.
8 0
3 years ago
Read 2 more answers
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