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marysya [2.9K]
2 years ago
12

Can someone plz help me with this one problem!!!!

Mathematics
2 answers:
Afina-wow [57]2 years ago
7 0

Answer:

no

Step-by-step explanation:

alexgriva [62]2 years ago
3 0
So 3 = x so you substitute it in the formula
y = 7(3)
y = 21
In the coordinates given y = 2 so it is incorrect
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A +-21 = -38<br><br>help me please​
ivann1987 [24]

Answer:

A= -17

Step-by-step explanation:

A +-21 = -38

    +21     +21

-38+21= -17

A= -17

5 0
2 years ago
Read 2 more answers
A policyholder has probability 0.7 of having no claims, 0.2 of having exactly one claim, and 0.1 of having exactly two claims. C
Sladkaya [172]

Answer:

0.89.

Step-by-step explanation:

So, we are given the following data or parameters or information which is going to aid in the solution to this question or problem. So, the data or parameters are;

(1). "probability 0.7 of having no claims, 0.2 of having exactly one claim, and 0.1 of having exactly two claims."

(2). " distributed on the interval [0,60] and are independent."

(3). "The insurer covers 100% of each claim."

So, taking (1) above we will have;

(Total probability= 0.7 + 0.2 + 0.1 = 0.1)

A = { (0.7 × 0)^2 + (0.2 × 1)^2 + ( 0.1 × 2)^2 =( 0.4) +( 0.4) = 0.8.

0.8 + total probability = 0.8 + 0.1 = 0.9.

So, from the question, we are to "Calculate the probability that the total benefit paid to the policyholder is 48 or less."

Hence, P< 48/A = P < 48/0.9 = P < 53.33.

Thus, 53.33/60 = 0.89.

6 0
3 years ago
In a clinical study, volunteers are tested for a gene that has been found to increase the risk for a disease. The probability th
MrRa [10]

Answer:

a) 0.984

b) 20 people

Step-by-step explanation:

a)

If The probability that a person carries the gene is 0.1, then in a sample of 20 people, 2 should carry the gene.

Now, we want to know how many samples there are with this property.

Since we have 20 elements where 18 are alike (do not carry the gene) and 2 are alike (carry the gene), we have to compute the number of permutations of 20 elements in which 18 are alike and 2 are alike. This number is

\frac{20!}{18!2!}=190

In this 190 20-tuples there are only 3 where the 2 carriers of the gene are in the first 3 places, namely

(1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)

(1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)

(0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)

where 1=carries the gene, 0=does not carry the gene.

So there are 190-3 = 187 elements in which the first 3 elements have no 2 carriers, hence the probability that 4 or more people will have to be tested until 2 of them with the gene are detected is 187/190 = 0.984 (98.4%) rounded to three decimal places.

b)

Given that the probability that a person carries the gene is 0.1, then in a sample of 20 people, 20*0.1 = 2 should carry the gene.

4 0
3 years ago
What is the area of the figure?
jok3333 [9.3K]
48.
For the rectangle, 4x6 =24
For the triangle, area of the triangle is 1/2 x base x height. So, 24.
Add together is 48
6 0
3 years ago
Which ordered pair is the solution to the system of equations? {x+3y=−12y=13x−6 (−5, −213) (−9, −9) (3, −5) (10, −223)
Sindrei [870]
I am not sure! What is the original equation?
7 0
3 years ago
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