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igomit [66]
1 year ago
6

a particular truck in a transportation fleet was driven 194,070 miles last year versus 329,919 miles this year . by what percent

did the annual distance driven increase
Mathematics
1 answer:
Ann [662]1 year ago
8 0

We take as the 100% the 194070 miles from last year. Then, we apply a rule of three:

\begin{gathered} 194070\rightarrow100\text{ \%} \\ 329919\rightarrow x\text{ \%} \\ \Rightarrow x=\frac{329919\cdot100}{194070}=\frac{32991900}{194070}=170 \\ x=170\text{ \%} \end{gathered}

Since 329,919 is the 170% of last year's miles, then the increase was of 70%

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Kayla needs $14,000 worth of new equipment for his shop. He can borrow this money at a discount rate of 10% for a year.
arlik [135]

Answer:

$15400

Step-by-step explanation:

Principle amount, P = $14000

Time, T = 1 year

Rate of interest, R = 10%

We know that maturity amount,

A = P\left (1+\frac{R}{100} \right )^{n}

where n is number of years

A = P\left (1+\frac{R}{100} \right )^{n}

A = 14000\left (1+\frac{10}{100}\right )^{1}

A = 14000\left (1+\frac{1}{10}\right )

A = 14000\left (\frac{11}{10}\right )

A = 15400

The maturity amount is $15400

6 0
3 years ago
What is the slope of (5,5) and (0,1)​
irina1246 [14]

Answer:

4/5

Step-by-step explanation:

Use the formulae

y2-y1 divided by x2- x1 then you will get your slope

3 0
3 years ago
There are more cups in 5 gallons than in 22 quarts
ololo11 [35]

Answer:

The answer is false

Step-by-step explanation:

22 quarts is 5.5 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
4 years ago
In a carnival drawing, a green ticket wins $1, a yellow ticket wins $5, and a blue ticket wins $10. There are 100 green tickets,
Rashid [163]

Answer:

D) 3:10

Step-by-step explanation:

Well only blues and yellows are at least $5. there are 25 yellows and 5 blues and if you add them together there's 30 in total. Now compare that 30 to 100 and you have 30:100. In the simplest form you have to divide both sides by 10 and it becomes 3:10.

3 0
3 years ago
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