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3 years ago

PLEASE IM BEGGING YOU

Mathematics

1 answer:

rodikova [14]3 years ago

5 0

Answer:b

Step-by-step explanation:

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Solve equation for x. State any restrictions. ax+2=bx+c

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Classify the equation, 7x+6=7x-5, as having one solution, no solution, or infinitely many solutions.

Ksenya-84 [330]

Answer:

Step-by-step explanation:

7x + 6 = 7x - 5

Collecting like terms

7x - 7x = -5 - 6

The x is canceled out so no solution

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3 years ago

7y+65.7=2 true or false(don't answer that I just don't want this taken down(unless you do actually have the answer) is anyone in

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8 0

3 years ago

Please help:((

kolezko [41]

Answer:

1/8 = 0.125

3/5= 0.60

30%=0.3

0.45=9/20

1.5=150%

Hey try to do these question by your self next time OwO

4 0

3 years ago

A stationary store has decided to accept a large shipment of ball-point pens if an inspection of 17 randomly selected pens yield

Leno4ka [110]

Answer:

0.762 = 76.2% probability that this shipment is accepted

Step-by-step explanation:

For each pen, there are only two possible outcomes. Either it is defective, or it is not. The probability of a pen being defective is independent from other pens. So we use the binomial probability distribution to solve this question.

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.

17 randomly selected pens

This means that $n = 17$

(a) Find the probability that this shipment is accepted if 10% of the total shipment is defective. (Use 3 decimal places.)

This is $P(X \leq 2)$ when $p = 0.1$ . So

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{17,0}.(0.1)^{0}.(0.9)^{17} = 0.167$

$P(X = 1) = C_{17,1}.(0.1)^{1}.(0.9)^{16} = 0.315$

$P(X = 2) = C_{17,2}.(0.1)^{2}.(0.9)^{15} = 0.280$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.167 + 0.315 + 0.280 = 0.762$

0.762 = 76.2% probability that this shipment is accepted

8 0

3 years ago