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3 years ago

Evaluate C=2 pi r for r=8 A. C=10 pi B. C=16 pi C. C=28 pi D. C=16

Mathematics

2 answers:

Verdich [7]3 years ago

5 0

Answer:C

Step-by-step explanation:Because I did this

Brrunno [24]3 years ago

4 0

Answer:

If `r` and `R` and the respective radii of the smaller and the bigger semi-circles then the area of the shaded portion in the given figure is: (FIGURE) `pir^2\ s qdotu n i t s` (b) `piR^2-pir^2\ s qdotu n i t s` (c) `piR^2+pir^2\ s qdotu n i t s` (d) `piR^2\ s qdotu n i t s`

Step-by-step explanation:

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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:

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

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.

5 0

3 years ago

What additional information is needed to prove triangle LMN is congruent to triangle KMN using the HL theorem?

IgorC [24]

<h2>Answer:</h2>

∠LMN is a right angle

<h2>Step-by-step explanation:</h2>

If we want to prove that two right triangles are congruent by knowing that the corresponding hypotenuses and one leg are congruent, we begin as follows:

Since two legs are congruent and we know this by the hash marks, then the triangle ΔLKN is isosceles.
By definition LN ≅ NK

If ∠LMN is a right angle, then MN is the altitude of triangle ΔLKN
Also MN is the bisector of LK, so KM ≅ ML
So we have two right triangles ΔLMN and ΔKM having the same lengths of corresponding sides

In conclusion, ΔLMN ≅ ΔKMN

3 0

3 years ago

Read 2 more answers

What two numbers add up to 18 and multiply to 80

mestny [16]

Answer:

8 and 10

Step-by-step explanation:

First, let's look at the factors of 80.

1 • 80 = 80

2 • 40 = 80

4 • 20 = 80

5 • 16 = 80

8 • 10 = 80

Look at each pair and evaluate their sum.

1 • 80 = 80

1 + 80 = 81

2 • 40 = 80

2 + 40 = 42

4 • 20 = 80

4 + 20 = 24

5 • 16 = 80

5 + 16 = 21

8 • 10 = 80

8 + 10 = 18

Our last pairing multiplies to 80 and adds up to 18. Therefore, your answer is 8 & 10.

Hope this helps!

4 0

2 years ago

ONLY NEED PART B HELP

bezimeni [28]

Answer: 48 model cars

Step-by-step explanation: 1. You know that the ratio of the number of model cars that Jim owns to the number of model cars Terrence owns is 4:3.

2. Let's call the number of model cars that Jim owns is , therefore, you can solve the exercise as following:

3. Therefore, Jims owns 48 model cars.

4 0

2 years ago

draw3 rows with 2 counters in each row write a word problem that can be acted out using these counters

Gnoma [55]

Pretend * is counter

* * *
* * *

the word problem could be in a box of chocolates there is 3 rows and 2 chocolates in each row how many chocolates are there total.

6 0

3 years ago

Read 2 more answers