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3 years ago

What is the mid point of AB

Mathematics

1 answer:

enot [183]3 years ago

3 0

Answer:

There's no specific answer

Step-by-step explanation:

If you want the Midpoint M between A and B then you should use the equation $\frac{A + B}{2}$

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95/8 -17/8 what does it equal

Rufina [12.5K]

Input the fraction = 95/8
Integer =11
Fraction part = 7/8
Answer = 11.875

8 0

2 years ago

Read 2 more answers

What is 5/11 minus negative 4/11

MA_775_DIABLO [31]

Answer: 1/11 and 0.09=0.10

Step-by-step explanation:

subtract by the numbers. subtract it's to take away one number from to another from the difference between the numbers.

5-4/11

5-4=1

=1/11

You can also round up to the nearest hundredths is 0.10.

Hope this helps!

Thanks!

Have a great day!

5 0

2 years ago

Read 2 more answers

Solve for x 4x – 4<8 AND 9x +5 > 23

Leno4ka [110]

Answer:

x<3 AND x>2

Step-by-step explanation:

First, 4x-4<8. Adding 4 to both sides, 4x<12, and dividing by 4, x<3.

Next, 9x+5>23. Subtracting 5 from both sides, 9x>18, and dividing by 9, x>2.

Therefore, x<3 AND x>2.

4 0

3 years ago

No question........................

frutty [35]

Answer:

Step-by-step explanation:

8 0

2 years ago

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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 0.0455 or 4.55%.

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