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2 years ago

What are the ratio of 1:6 and 3:8, 2:14and 3:42

Mathematics

1 answer:

GrogVix [38]2 years ago

3 0

The ratio of 1:6 is 1/6

The ratio of 3:8 is 3/8

The ratio of 2:14 is 1/7

The ratio of 3:42 is 1/14

<em>Please mark brainliest if that helped!</em>

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Find the measure of arc DF (with an explanation of how to solve please!)

allochka39001 [22]

Answer:

mDF = 90°

(Assuming arc mCE is 5x + 10)

Step-by-step explanation:

When we have two chords crossing in a circle, there is a property where the angle in the cross point is equal half the sum of the corresponding arcs.

So in this case, we have:

70° = (mCE + mDF)/2

Assuming mCE is 5x + 10, we have:

70 = (5x + 10 + 11x + 2)/2

70 = (16x + 12)/2

70 = 8x + 6

8x = 64

x = 8

So the arc mDF is:

mDF = 11x + 2 = 11 * 8 + 2 = 90°

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3 years ago

Simple interest 34,100 at 4% for 3 years

Kaylis [27]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

21 = − 6 x − 3 x + 4 + 4 x

horsena [70]

Answer:

x=-17/5

Step-by-step explanation:

1. Simplify -6x-3x+4+4x−6x−3x+4+4x to -5x+4−5x+4.

21=−5x+4

2. Subtract 44 from both sides.

21−4=−5x

3. Simplify 21-4 to 17.

17=-5x

4. Divided both sides by -5.

-17/5=x

5. Switch sides.

x=-17/5

Hope this helps.

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3 years ago

Please answer these 2 questions

Mamont248 [21]

Answer:

FINE ITS A AND B

Step-by-step explanation:

3 0

3 years ago

A research study uses 800 men under the age of 55. Suppose that 30% carry a marker on the male chromosome that indicates an incr

ss7ja [257]

Answer:

a) There is a 12.11% probability that exactly 1 man has the marker.

b) There is a 85.07% probability that more than 1 has the marker.

Step-by-step explanation:

There are only two possible outcomes: Either the men has the chromosome, or he hasn't. So we use the binomial probability distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

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 $\pi$ is the probability of X happening.

In this problem, we have that:

30% carry a marker on the male chromosome that indicates an increased risk for high blood pressure, so $\pi = 0.30$

(a) If 10 men are selected randomly and tested for the marker, what is the probability that exactly 1 man has the marker?

10 men, so $n = 10$

We want to find P(X = 1). So:

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

$P(X = 1) = C_{10,1}.(0.30)^{1}.(0.7)^{9} = 0.1211$

There is a 12.11% probability that exactly 1 man has the marker.

(b) If 10 men are selected randomly and tested for the marker, what is the probability that more than 1 has the marker?

That is $P(X > 1)$

We have that:

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

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

We also have that:

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

$P(X = 0) = C_{10,0}.(0.30)^{0}.(0.7)^{10} = 0.0282$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0282 + 0.1211 = 0.1493$

Finally

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.1493 = 0.8507$

There is a 85.07% probability that more than 1 has the marker.

3 0

3 years ago