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2 years ago

I need helppppppppppppp

Mathematics

1 answer:

ohaa [14]2 years ago

5 0

Simplify

4ax+12-3ax=25+3a

ax+12=25+3a

ax-3a=25-12

a(x-3)=13

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At store A, rulers are sold individually. The cost y of x rulers is represented by the equation y = 0.95x. The costs of rulers a

makkiz [27]

Answer:

The ruler at store A cost more

Step-by-step explanation:

Ruler A=$0.95 per ruler

Ruler B=$0.92 per ruler

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3 years ago

9. What’s the answer to this question!

spin [16.1K]

Answer:

the answer is b

Step-by-step explanation:

first you simplify the numbers by 25 and end up with x - 5 over x^2 and then you simplify by x and you get -5 over x

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2 years ago

What is the probability that you would obtain a sum of 7 or a sum of 11 on the first roll? (Rolling two dice.)

crimeas [40]

Answer:

2/9

Step-by-step explanation:

Total outcome = 36

Sum of 7 = 6

Prob of sum of 7 = 6/36

Sum of 11 = 2

Prob of sum of 11 = 2/36

Prob of sum of 7 or 11 = 6/36+2/36

Prob of sum of 7 or 11 = (6+2)/36

Prob of sum of 7 or 11 = 8/36 = 2/9

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3 years ago

If g is the variable, which mathematical sentence expresses the information below? "The number of gallons of gas in the tank plu

Firdavs [7]

An equation for the sentence would be g + 7 = 12, where g is the unknown number of gallons.

5 0

3 years ago

Read 2 more answers

A ferry will safely accommodate 82 tons of passenger cars. Assume that the meanweight of a passenger car is 2 tons with standard

Shalnov [3]

Answer:

The probability that the maximum safe-weight will be exceeded is <u>0.0455 or 4.55%</u>.

Step-by-step explanation:

Given:

Maximum safe-weight of 37 cars = 82 tons

∴ Maximum safe-weight of 1 car (x) = 82 ÷ 37 = 2.22 tons (Unitary method)

Mean weight of 1 car (μ) = 2 tons

Standard deviation of 37 cars = 0.8 tons

So, standard deviation of 1 car is given as:

$\sigma=\frac{0.8}{\sqrt{37}}=0.13$

Probability that maximum safe-weight is exceeded, P(x > 2.22) = ?

The sample is normally distributed (Assume)

Now, let us determine the z-score of the mean weight.

The z-score is given as:

$z=\frac{x-\mu}{\sigma}\\\\z=\frac{2.22-2}{0.13}\\\\z=\frac{0.22}{0.13}=1.69$

Now, finding P(x > 2.22) is same as finding P(z > 1.69).

From the z-score table of normal distribution curve, the value of area under the curve for z < 1.69 is 0.9545.

But we need the area under the curve for z > 1.69.

So, we subtract from the total area. Total area is 1 or 100%.

So, $P(z > 1.69) = 1 - P(z < 1.69)$

$P(z>1.69)=1-0.9545=0.0455\ or\ 4.55\%$

Therefore, the probability that the maximum safe-weight will be exceeded is 0.0455 or 4.55%.

8 0

3 years ago