1. In the following triangle, A = 62, 6 = 96, and a = 120. Use the Law of Sines to find the

It would be: 30/75 = 6/15 = 2/5

So, it's reduced & most simple form is 2/5

Hope this helps!

8 0

3 years ago

Please help me Create a system of two equations but do not solve the system please actually help me please and thank you

mel-nik [20]

Answer:

7r + g = 46

7r + 2g = 50

Step-by-step explanation:

Hi there!

Matt and Trevon both bought supplies from a store, and we know that Matt bought 7 rose bushes and 1 geranium, and spent a total of $46 while Trevon bought 7 rose bushes and 2 geraniums, and spent a total of $50

We know that each rose bush costs a certain amount, and each geranium must cost a certain amount, but we don't know what that amount is; it's also possible that the amounts could be different

So let's say that the cost of one rose bush is r dollars; for Matt, that means that he spent 7r dollars on 7 rose bushes. Trevon also bought 7 rose bushes, meaning he also spent 7r dollars on rose bushes.

Since the cost of the geraniums can be different from the cost of every rose bush, let's express it with a different variable, like g.
Since Matt bought one geranium, he spent g dollars on that one geranium.
Trevon bought 2 geraniums, so he spent 2g dollars on the geraniums

The total cost of what Matt spent on flowers was 7r + g dollars (7 rose bushes + 1 geranium), which is equal to 46 dollars

As an equation, that can be written as 7r + g = 46

Trevon spent 7r + 2g dollars (7 rose bushes + 2 geraniums), which gave him a total of $50

As an equation, that can be written as 7r + 2g = 50
The two equations therefore are:
7r + g =46
7r+2g=50
Hope this helps!

Take a look here to practice with a similar problem if you wish (you don't have to look at the actual solving portion): brainly.com/question/14033905

8 0

2 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:

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

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

So

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

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:

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