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3 years ago

(35+27)÷4×12= what??

2 answers:

8 0

We have to do BIMDAS

so starting with the brackets we do

35+27

which is 62

62÷4×12

then we work left to right for multiplication and division

62÷4

which is 15.5

then we finish with 15.5×12

which is 186

hope it helps.

Stels [109]3 years ago

3 0

(35+27)÷4×12
62÷4×12
15.5×12
186

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3 years ago

Among persons donating blood to a clinic, 85% have Rh+ blood (that is, the Rhesus factor is present in their blood.) Six people

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Answer:

a) There is a 62.29% probability that at least one of the five does not have the Rh factor.

b) There is a 22.36% probability that at most four of the six have Rh+ blood.

c) There need to be at least 8 people to have the probability of obtaining blood from at least six Rh+ donors over 0.95.

Step-by-step explanation:

For each person donating blood, there are only two possible outcomes. Either they have Rh+ blood, or they do not. This means that we use the binomial probability distribution to solve this problem.

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 combinatios 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.

In this problem we have that:

$p = 0.85, n = 6$ .

a) fine the probability that at least one of the five does not have the Rh factor.

Either all six have the factor, or at least one of them do not. The sum of the probabilities of these events is decimal 1. So:

$P(X < 6) + P(X = 6) = 1$

$P(X < 6) = 1 - P(X = 6)$

In which:

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

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

$P(X < 6) = 1 - P(X = 6) = 1 - 0.3771 = 0.6229$

There is a 62.29% probability that at least one of the five does not have the Rh factor.

b) find the probability that at most four of the six have Rh+ blood.

Either more than four have Rh+ blood, or at most four have. So

$P(X \leq 4) + P(X > 4) = 1$

$P(X \leq 4) = 1 - P(X > 4)$

In which

$P(X > 4) = P(X = 5) + P(X = 6)$

$P(X = 5) = C_{6,5}.(0.85)^{5}.(0.15)^{1} = 0.3993$

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

$P(X > 4) = P(X = 5) + P(X = 6) = 0.3993 + 0.3771 = 0.7764$

$P(X \leq 4) = 1 - P(X > 4) = 1 - 0.7764 = 0.2236$

There is a 22.36% probability that at most four of the six have Rh+ blood.

c) The clinic needs six Rh+ donors on a certain day. How many people must donate blood to have the probability of obtaining blood from at least six Rh+ donors over 0.95?

With 6 donors:

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

37.71% probability of obtaining blood from at least six Rh+ donors over 0.95.

With 7 donors:

$P(X = 6) = C_{7,6}.(0.85)^{6}.(0.15)^{1} = 0.3960$

0.3771 + 0.3960 = 0.7764 = 77.64% probability of obtaining blood from at least six Rh+ donors over 0.95.

With 8 donors

$P(X = 6) = C_{8,6}.(0.85)^{6}.(0.15)^{2} = 0.2376$

0.3771 + 0.3960 + 0.2376 = 1.01 = 101% probability of obtaining blood from at least six Rh+ donors over 0.95.

There need to be at least 8 people to have the probability of obtaining blood from at least six Rh+ donors over 0.95.

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3 years ago

-1/3(6x-15)+6x>8x-11 solve.

arsen [322]

In this question, you're solving the inequality for x.

Solve:

-1/3(6x - 15) + 6x > 8x - 11

Distribute the -1/3 to the variables in the parenthesis.

-2x + 5 + 6x > 8x - 11

Combine like terms

4x + 5 > 8x - 11

Subtract 8x from both sides

-4x + 5 > - 11

Subtract 5 from both sides

-4x > -16

Divide both sides by -4, while also flipping the inequality, since you're dividing by a negative.

x < 4

Answer:

x < 4

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3 years ago

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Answer:

Step-by-step explanation:

Given

2x² + x - 1 = 2 ( subtract 2 from both sides )

2x² + x - 3 = 0

Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term.

product = 2 × - 3 = - 6 and sum = + 1

The factors are - 2 and + 3

Use these factors to split the x- term

2x² - 2x + 3x - 3 = 0 ( factor the first/second and third/fourth terms )

2x(x - 1) + 3(x - 1) = 0 ← factor out (x - 1) from each term

(x - 1)(2x + 3) = 0

Equate each factor to zero and solve for x

x - 1 = 0 ⇒ x = 1

2x + 3 = 0 ⇒ 2x = - 3 ⇒ x = - $\frac{3}{2}$

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