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3 years ago

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Mathematics

1 answer:

finlep [7]3 years ago

7 0

Answer:

Hi what language is that I don't know that language so I cannot answer your question

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Math question about percent error. Picture of the problem attached. PLease hurry.

kifflom [539]

I think it is C. but Im not positive

4 0

3 years ago

Read 2 more answers

Bad gums may mean a bad heart. Researchers discovered that 79% of people who have suffered a heart attack had periodontal diseas

zysi [14]

Answer:

(A) 0.297

(B) 0.595

Step-by-step explanation:

Let,

H = a person who suffered from a heart attack

G = a person has the periodontal disease.

Given:

P (G|H) = 0.79, P(G|H') = 0.33 and P (H) = 0.15

Compute the probability that a person has the periodontal disease as follows:

$P(G)=P(G|H)P(H)+P(G|H')P(H')\\=P(G|H)P(H)+P(G|H')(1-P(H))\\=(0.79\times0.15)+(0.33\times(1-0.15))\\=0.399$

(A)

The probability that a person had periodontal disease, what is the probability that he or she will have a heart attack is:

$P(H|G)=\frac{P(G|H)P(H)}{P(G)}\\=\frac{0.79\times0.15}{0.399} \\=0.29699\\\approx0.297$

Thus, the probability that a person had periodontal disease, what is the probability that he or she will have a heart attack is 0.297.

(B)

Now if the probability of a person having a heart attack is, P (H) = 0.38.

Compute the probability that a person has the periodontal disease as follows:

$P(G)=P(G|H)P(H)+P(G|H')P(H')\\=P(G|H)P(H)+P(G|H')(1-P(H))\\=(0.79\times0.38)+(0.33\times(1-0.38))\\=0.5048$

Compute the probability of a person having a heart attack given that he or she has the disease:

$P(H|G)=\frac{P(G|H)P(H)}{P(G)}\\=\frac{0.79\times0.38}{0.5048}\\ =0.59469\\\approx0.595$

The probability of a person having a heart attack given that he or she has the disease is 0.595.

4 0

3 years ago

Find the slope by changing form Standard Form to Slope Intercept Form 7x- 8y = -56

Sholpan [36]

Answer:

y = 7/8x + 7

Step-by-step explanation:

7x- 8y = -56

-7x -7x

-8y = -7x - 56

y = 7/8x + 7

6 0

2 years ago

Read 2 more answers

A 41 gram sample of a substance that's used to detect explosives has a k-value of 0.1392. Find the substance's half life, in day

-BARSIC- [3]

Answer:

The substance half-life is of 4.98 days.

Step-by-step explanation:

Equation for an amount of a decaying substance:

The equation for the amount of a substance that decay exponentially has the following format:

$A(t) = A(0)e^{-kt}$

In which k is the decay rate, as a decimal.

k-value of 0.1392.

This means that:

$A(t) = A(0)e^{-0.1392t}$

Find the substance's half life, in days.

This is t for which $A(t) = 0.5A(0)$ . So

$A(t) = A(0)e^{-0.1392t}$

$0.5A(0) = A(0)e^{-0.1392t}$

$e^{-0.1392t} = 0.5$

$\ln{e^{-0.1392t}} = \ln{0.5}$

$-0.1392t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.1392}$

$t = 4.98$

The substance half-life is of 4.98 days.

5 0

3 years ago

One number is 6 more than a second number, the sum of the two numbers is 330. what are the numbers?

maksim [4K]

Let's have the first number, the larger number, be x. We'll have the second, smaller number be y.

We know that x = y + 6, since x is 6 greater than y.

We also know that 330 = x + y.

Because x = y + 6, 330 = y + 6 + y, which simplifies to 330 = 2y + 6.

Now all we need to do is simplify the equation. First, we subtract 6 from both sides:

330 - 6 = 324

2y + 6 - 6 = 2y.

So we have 324 = 2y. Then we divide both sides by 2 to get:

162 = y

Plug in y = 162 into the equation x = y + 6 to get:

x = 162 + 6

x = 168

Let's check to make sure our answer is right. 168 is 6 more than 162. 162 + 168 equals 330. So our two numbers are 168 and 162.

7 0

3 years ago