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lions [1.4K]
3 years ago
6

GIVING BRAINLIEST i know it aint math but no one does the buisness subject

Mathematics
2 answers:
Setler79 [48]3 years ago
5 0

Answer:

Post secondary

Step-by-step explanation:

Sorr if itswrong

snow_lady [41]3 years ago
3 0

Answer:

           

Step-by-step explanation:

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30+6⋅7+8 need help asap!!!!
Sladkaya [172]

Answer:

80 :D!

Step-by-step explanation:

30 + (6 x 7) + 8

30 + 42 + 8

72 + 8

= 80

Hope this helps! Brainliest plz?

8 0
2 years ago
Read 2 more answers
-7(n - 2) + 2n = -5(n + 6)
skad [1K]

Given the expression below

-7(n-2)+2n=-5(n+6)

To find n

Open the brackets

\begin{gathered} -7(n-2)+2n=-5(n+6) \\ -7n+14+2n=-5n-30 \\ -7n+2n+14=-5n-30 \\ -5n+14=-5n-30 \end{gathered}

Collect like terms

\begin{gathered} -5n+14=-5n-30 \\ -5n-(-5n)=-30-14 \\ 0\ne-44 \end{gathered}

Since, the sides are not equal,

Hence, there is no solution

7 0
1 year ago
Can you please Help me with this?<br><br> (Greatly Appreciated)
exis [7]

Answer:

x=4²-((3×2)÷2)

x=13

Step-by-step explanation:

assuming this is a right triangle (90dgr)

area of the square... a×a

area of the right triangle..= (width×height) ÷2

3 0
3 years ago
Can anyone help me on this ?
andrew-mc [135]

Answer:

37.9

Step-by-step explanation:

21.600/21.6+16.3=37.9

8 0
2 years ago
The automatic opening device of a military cargo parachute has been designed to open when the parachute is 185 m above the groun
agasfer [191]

Answer:

0.0193 = 1.93% probability that there is equipment damage to the payload of at least one of five independently dropped parachutes.

Step-by-step explanation:

For each parachute, there are only two possible outcomes. Either there is damage, or there is not. The probability of there being damage on a parachute is independent of any other parachute, which means that the binomial probability distribution is used to solve this question.

To find the probability of damage on a parachute, the normal distribution is used.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Probability of a parachute having damage.

The opening altitude actually has a normal distribution with mean value 185 and standard deviation 32 m, which means that \mu = 185, \sigma = 32

Equipment damage will occur if the parachute opens at an altitude of less than 100 m, which means that the probability of damage is the p-value of  Z when X = 100. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{100 - 185}{32}

Z = -2.66

Z = -2.66 has a p-value of 0.0039.

What is the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes?

0.0039 probability of a parachute having damage, which means that p = 0.0039

5 parachutes, which means that n = 5

This probability is:

P(X \geq 1) = 1 - P(X = 0)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{5,0}.(0.0039)^{0}.(0.9961)^{5} = 0.9807

Then

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.9807 = 0.0193

0.0193 = 1.93% probability that there is equipment damage to the payload of at least one of five independently dropped parachutes.

5 0
3 years ago
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