All categories

3 years ago

What is the probability that a person who is above 35 years old has a hemoglobin level of 9 or above?

Mathematics

2 answers:

lorasvet [3.4K]3 years ago

4 0

Answer:

Option C. $P=0.531$

Step-by-step explanation:

we know that

The probability of an event is the ratio of the size of the event space to the size of the sample space.

The size of the sample space is the total number of possible outcomes

The event space is the number of outcomes in the event you are interested in.

Let

x------> size of the event space (total person's hemoglobin level of 9 or above with age above 35 years)

y-----> size of the sample space (total person's age above 35 years)

$P=\frac{x}{y}$

In this problem we have

$y=162$

Complete the table to find the total person's hemoglobin level of 9 or above (person's age above 35)

Let

y------> total person's hemoglobin level between 9 and 11 (person's age above 35)

$y+76+40=162$

$y=46$

Find the value of x

$x=46+40=86$

substitute the values

$P=\frac{86}{162}=0.531$

earnstyle [38]3 years ago

3 0

Answer:

The answer is C. 0.531

Step-by-step explanation:

Because

76+x+40= 162.

x= 162-76-40

x= 46

Hence,

40+46=86.

Therefore, probability that a person who is above 35 years old has a hemoglobin level of 9 or above is,

86/162= 0.531.

You might be interested in

a farmer bought a number of pigs for $180. However, 3 of them died before he could sell the rest at a profit of 8 per pig. His t

Gre4nikov [31]

Answer:

He bought approximately 16 pigs.

Step-by-step explanation:

In order to find how many pigs the farmer had originally all you have to do is add three to amount of pigs he sold, since three of the died. In order to find how many he sold all you need to do is divide 109 by 8, since he made a profit of $8 per pig. 109 / 8 = 13.625, 13.625 + 3 = 16.625

4 0

3 years ago

How do i solve 4m^2+3m-1=0 using the quadratic formula

elixir [45]

Quadratic formulat is used when a^2+bx+c=0 x= $\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}$ 4m^2+3m-1=0 a=4 b=3 c=-1 input x= $\frac{-3+/- \sqrt{3^{2}-4(4)(-1)} }{2(4)}$ x= $\frac{-3+/- \sqrt{9+16} }{8}$ x= $\frac{-3+/- \sqrt{25} }{8}$ x= $\frac{-3+/-5 }{8}$ x= $\frac{-8}{8}$ or $\frac{2}{8}$ x=-1 or 1/4

8 0

3 years ago

What is StartFraction negative 23 over 6 EndFraction minus StartFraction 7 over 3 EndFraction

creativ13 [48]

Answer: $\frac{16}{3}$

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

4m-m-1/2=3m+1/3 what is the value of m

sammy [17]

Answer:

Correct option is

1,−1/2

For equal roots D=b

−4ac=0

⇒[2(m+1)]

−4m(3m+1)=0

⇒m

+2m+1−3m

−m=0

⇒−2m

+m+1=0

⇒2m

−m−1=0⇒(m−1)(m+

)=0

⇒m=1,

−1

7 0

2 years ago

The length of pregnancy isn’t always the same. In pigs, the length of pregnancies varies according to a normal distribution with

Nesterboy [21]

a. percent of pig pregnancies that are longer than 106 days

Since we have a normal distribution here and the average number of days is 106, we can say that 50% of the pig pregnancies are longer than 106 days.

b. percent of pig pregnancies that are shorter than 111 days

To get the percentage, we will have to convert x = 111 days into a z-score first. The formula is:

$z=\frac{x-\mu}{\sigma}$

where x = raw data, μ = population mean, and σ = population SD.

Since these 3 pieces of information are already given in the question, let's plug them into the equation above.

$z=\frac{111-106}{5}=\frac{5}{5}=1\sigma$

Therefore, 111 days is located 1 standard deviation to the right of the mean.

To find the percentage of pig pregnancies shorter than 111 days, we need to find the area covered to the left of 1 SD.

To find the area covered to the left of 1SD, we need to use the standard normal distribution table.

Based on the table, the area covered to the left of 1SD is 0.8413. Multiplying the area by 100, we get 84.13. Therefore, 84.13% of the pig pregnancies are shorter than 111 days.

5 0

1 year ago