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alexandr1967 [171]
3 years ago
15

Gina has a spinner with five equal parts numbered one thru five. If Gina spins the spinner one time, what is the probability tha

t the spinner will land on the number three?
Mathematics
1 answer:
Alex787 [66]3 years ago
3 0

Answer:

1/5

Step-by-step explanation:

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Please help ASAP: Factorise the following
Alex

Answer:

Step-by-step explanation:

Slide 1:

A) (x+2)(x+8)

B) (x+3)(x+4)

C) (x+1)(x+12)

Slide 2:

A) (x-1)(x+4)

B) (x+1)(x-3)

C) (x-2)(x+4)

Hope this helps!

5 0
3 years ago
Tay–Sachs Disease Tay–Sachs disease is a genetic disorder that is usually fatal in young children. If both parents are carri
BARSIC [14]

Answer:

a) 0.0156 = 1.56% probability that all children will develop the disease.

b) 0.4219 = 42.19% probability that only one child will develop the disease.

c) 0.1406 = 14.06% probability that the third children will develop the disease, given that the first two did not.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they carry the disease, or they do not. The probability of a children carrying the disease is independent of any other children, which means that the binomial probability distribution is used to solve this question.

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 combinations 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.

The probability that their offspring will develop the disease is approximately .25.

This means that p = 0.25

Three children:

This means that n = 3

Question a:

This is P(X = 3). So

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

P(X = 3) = C_{3,3}.(0.25)^{3}.(0.75)^{0} = 0.0156

0.0156 = 1.56% probability that all children will develop the disease.

Question b:

This is P(X = 1). So

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

P(X = 1) = C_{3,1}.(0.25)^{1}.(0.75)^{2} = 0.4219

0.4219 = 42.19% probability that only one child will develop the disease.

c. The third child will develop Tay–Sachs disease, given that the first two did not.

Third independent of the first two, so just multiply the probabilities.

First two do not develop, each with 0.75 probability.

Third develops, which 0.25 probability. So

p = 0.75*0.75*0.25 = 0.1406

0.1406 = 14.06% probability that the third children will develop the disease, given that the first two did not.

6 0
3 years ago
A triangle is placed in a semicircle with a radius of 5 yd, as shown below. Find the area of the shaded region.
Svetradugi [14.3K]

Answer:

  14.25 yd^2

Step-by-step explanation:

The triangle area is given by the formula ...

  A = 1/2bh

Here, the base is 10 yd, and the height is 5 yd, so ...

  triangle area = (1/2)(10 yd)(5 yd) = (5 yd)^2 = 25 yd^2

__

The semicircle is half the area of a circle with radius 5 yd, so its area is ...

  semicircle area = (1/2)π(5 yd)^2 = 12.5π yd^2 ≈ 39.25 yd^2

__

The shaded region is the difference between these areas:

  shaded area = 39.25 yd^2 -25 yd^2 = 14.25 yd^2

7 0
3 years ago
How to show my work when I have a math problem
Over [174]

Answer:

Here is how

Step-by-step explanation:

You find your answer to the problem and explain how do did it or you write the steps of how you did the problem. Also put your answer.

5 0
3 years ago
A card is selected randomly from a jar that contains 15 cards, numbered from 25 to 39. What is the probability that the card sel
VikaD [51]

Answer:

The probability that the card selected bears a number less than 34 is 0.3333.

Step-by-step explanation:

Let random variable <em>X</em> be defined as the number on the selected card.

There are <em>N</em> = 15 total cards.

The number on the cards are as follows:

S = {25, 26, 27,..., 38, 39}

The probability of an event, <em>E</em> is the ratio of the number of favorable outcomes to the total number of outcomes.

P(E)=\frac{n(E)}{N}

In this case we need to compute the probability that the card selected bears a number less than 34.

The favorable outcomes are:

<em>s</em> = {25, 36, 37, 38, 39}

<em>n</em> (X < 34) = 5

Compute the probability that the card selected bears a number less than 34 as follows:

P(X

                  =\frac{5}{15}\\\\=\frac{1}{3}\\\\=0.3333

Thus, the probability that the card selected bears a number less than 34 is 0.3333.

7 0
3 years ago
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