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2 years ago

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tigry1 [53]2 years ago

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A car travels 15 miles in 15 minutes how many miles does it travel in 2 hours

zubka84 [21]

Answer:

120 miles in 2 hour

Step-by-step explanation:

One hour has 4 times 15 minutes, so you need to do 15 miles x 4 and that time 2 because its 2 hours.

5 0

3 years ago

IfA=(-2,-4) and B=(-8,4), what is the length of AB?

lisabon 2012 [21]

Answer:

<h2>10 units </h2>

solution,

A(-2,-4)--->(X1,y1)

B(-8,4)----->(x2,y2)

now,

$ab = \sqrt{(x2 - x1) ^{2} + {(y2 - y1)}^{2} } \\ \: \: \: \: = \sqrt{ (- 8 - ( - 2)) ^{2} + (4 - ( - 4)) ^{2} } \\ \: \: \: = \sqrt{( - 8 + 2) ^{2} + {(4 + 4)}^{2} } \\ \: \: \: = \sqrt{( - 6) ^{2} + {(8)}^{2} } \\ \: \: = \sqrt{36 + 64} \\ \: \: \: = \sqrt{100} \\ \: \: \: = \sqrt{ {(10)}^{2} } \\ \: \: \: \: = 10 \: units$

Hope this helps..

Good luck on your assignment

3 0

3 years ago

The original price of a video game is $45. What is the sale price after a 25% discount.

Ghella [55]

Answer:

33.75

Step-by-step explanation:

5 0

3 years ago

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

Let M = {2, 5, 7} and N = {4, 5, 8}.

levacccp [35]

C is the answer of the questions

5 0

3 years ago