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3 years ago

8

Planes flies 2951 miles in 6.5 hours what is the speed of the airplane miles per hour

2 answers:

aleksandr82 [10.1K]3 years ago

5 0

2951mi/6.5 h = 454 mph

pantera1 [17]3 years ago

3 0

Divide 2951 by 6.5 to get the answer

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Solve the inequality.<br> 4│x + 5│ - 2 ≤ 10

castortr0y [4]

Step-by-step explanation: To solve this absolute value inequality,

our goal is to get the absolute value by itself on one side of the inequality.

So start by adding 2 to both sides and we have 4|x + 5| ≤ 12.

Now divide both sides by 3 and we have |x + 5| ≤ 3.

Now the the absolute value is isolated, we can split this up.

The first inequality will look exactly like the one

we have right now except for the absolute value.

For the second one, we flip the sign and change the 3 to a negative.

So we have x + 5 ≤ 3 or x + 5 ≥ -3.

Solving each inequality from here, we have x ≤ -2 or x ≥ -8.

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2 years ago

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Find the product: (-3)(-5)

MariettaO [177]

Answer:

15

Step-by-step explanation:

-3 times -5 equals 15 because two negatives multiplied by each other equals a positive.

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3 years ago

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Which is greater, 300 kelvins or 25 degrees Celsius? Show your work

Mademuasel [1]

300 kelvin is greater.

300 kelvin equals approximately 26.85°C

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3 years ago

In the given figure △ABC ≅△DEC. Which of the following relations can be proven using CPCTC ?

Serhud [2]

Option B:

$\overline{A B}=\overline{D E}$

Solution:

In the given figure $\triangle A B C \cong \triangle D E C$ .

If two triangles are similar, then their corresponding sides and angles are equal.

By CPCTC, in $\triangle A B C \ \text{and}\ \triangle D E C$ ,

$\overline{AB }=\overline{DE}$ – – – – (1)

$\overline{B C}=\overline{EC}$ – – – – (2)

$\overline{ CA}=\overline{CD}$ – – – – (3)

$\angle ACB=\angle DCE$ – – – – (4)

$\angle ABC=\angle DEC$ – – – – (5)

$\angle BAC=\angle EDC$ – – – – (6)

Option A: $\overline{B C}=\overline{D C}$

By CPCTC proved in equation (2) $\overline{B C}=\overline{EC}$ .

Therefore $\overline{B C}\neq \overline{D C}$ . Option A is false.

Option B: $\overline{A B}=\overline{D E}$

By CPCTC proved in equation (1) $\overline{AB }=\overline{DE}$ .

Therefore Option B is true.

Option C: $\angle A C B=\angle D E C$

By CPCTC proved in equation (4) $\angle ACB=\angle DCE$ .

Therefore $\angle A C B\neq \angle D E C$ . Option C is false.

Option D: $\angle A B C=\angle E D C$

By CPCTC proved in equation (5) $\angle ABC=\angle DEC$ .

Therefore $\angle A B C\neq \angle E D C$ . Option D is false.

Hence Option B is the correct answer.

$\Rightarrow\overline{A B}=\overline{D E}$

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2 years ago

The computer that controls a bank's automatic teller machine crashes a mean of 0.6 times per day. What is the probability that,

professor190 [17]

Answer:

0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Mean of 0.6 times a day

7 day week, so $\mu = 7*0.6 = 4.2$

What is the probability that, in any seven-day week, the computer will crash less than 3 times? Round your answer to four decimal places.

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-4.2}*(4.2)^{0}}{(0)!} = 0.0150$

$P(X = 1) = \frac{e^{-4.2}*(4.2)^{1}}{(1)!} = 0.0630$

$P(X = 2) = \frac{e^{-4.2}*(4.2)^{2}}{(2)!} = 0.1323$

So

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0150 + 0.0630 + 0.1323 = 0.2103$

0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.

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2 years ago