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3 years ago

Caroline drove 350 miles to her grandmothers house. The trip took her 5 1/4 hours what was her average speed in miler per hour?

Mathematics

1 answer:

SpyIntel [72]3 years ago

8 0

The answer is 66.6 repeated so if you round 66.6 repeated you get 67 so I think the answer is about 67 miles per hour

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Please solve<br>x^2+x-6006=0

Ratling [72]

Step-by-step explanation:

x^2+x-6006=0

x^2+78x-77x-6006=0

x(x+78)-77(x+78)=0

(x-77)(x+78)=0

=>x=77 or x=78

7 0

3 years ago

I need answers for A -F

mestny [16]

Answer:

A. 103º

B. 96º

C. m<4 = m<10

D. m<1 = m<16

E. m<5 = m<13

F. m<14 + m<15 = 180º

(Not entirely sure for C-F... But I hope this helps!)

8 0

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

Leona [35]

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:

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