Ask question

Ask question

All categories

2 years ago

3

sam buys milk from a different shop. he writes an equation for the amount he spends on milk, p=3.4g, where p is the cost and g i

s the gallons of milk brought. Who buys milk at a lower rate, and what is the price?

1 answer:

Talja [164]2 years ago

6 0

What is the first part of the questions??

You might be interested in

According to a study in a medical journal, 202 of a sample of 5,990 middle-aged men had developed diabetes. It also found that m

tekilochka [14]

Answer:

0.0588 = 5.88% probability that a middle-aged man with diabetes is very active

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Has diabetes.

Event B: Is very active.

Probability of having diabetes:

To find this probability, we take in consideration that:

It also found that men who were very active (burning about 3,500 calories daily) were a fourth as likely to develop diabetes compared with men who were sedentary. Assume that one-fifth of all middle-aged men are very active, and the rest are classified as sedentary.

So the probability of developing diabetes is:

x of 4/5 = x of 0.8(not active)

x/4 = 0.25x of 1/5 = 0.2(very active). So

$P(A) = 0.8x + 0.25*0.2x = 0.85x$

Probability of developing diabetes while being very active:

0.25x of 0.2. So

$P(A \cap B) = 0.25x*0.2 = 0.05x$

What is the probability that a middle-aged man with diabetes is very active?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.05x}{0.85x} = \frac{0.05}{0.85} = 0.0588$

0.0588 = 5.88% probability that a middle-aged man with diabetes is very active

4 0

2 years ago

Supervisor: "You took 500 calls in 30 days." Representative: "That means I took an average of __________ calls per day. I'm pret

ollegr [7]

Answer:

17 calls per day

Step-by-step explanation:

500 calls ÷ 30 days= 16.6666667 calls per day

<u>OR</u>

≈ 17 calls per day

4 0

2 years ago

Reconsider the system defect situation described in Exercise 26 (Section 2.2). a. Given that the system has a type 1 defect, wha

nexus9112 [7]

Answer:

hello exercise 26 is missing attached below is exercise 26

answer: a) 0.5 (b) 0.833 (c) 0.357 (d) 0.833

Step-by-step explanation:

A1... A3 = represents defects in system 1 to 3

A' 1 ---- A'3 = represents no defects in systems 1 to 3

P( A1 ∩ A2 ) = 0.06 ( as calculated )

P ( A1 ∩ A3 ) = 0.03 ( as calculated )

P ( A2 ∩ A3 ) = 0.02 ( as calculated )

a) Probability of having a type 2 defect

The probability of A given system having both Type 2 defect given that it has a type 1 defect is considered conditional probability

= p ( A2 | A 1 ) = $\frac{P(A1 n A2)}{P(A1)}$

= $\frac{0.06}{0.12}$ = 0.5

B) Probability of having all three defects given that it has type 1 defect

= P ( A1 ∩ A2 ∩ A3 | A1 ) = $\frac{P( A1 n A2 n A3 )}{P(A1)}$

= $\frac{0.01}{0.12}$ = 0.0833 ≈ 0.8

C) probability of having exactly one type of defect given that it has atleast one type of defect

= P ( exactly one Ai | at least one Ai ) = 0.357

attached below is the detailed solution

D) probability of not having the third defect given that it has the first two types of defects

= P ( A'3 | A1 ∩ A2 ) = 0.833

attached below is the detailed solution

8 0

2 years ago

On a coordinate plane, 2 triangles are shown. The first triangle has points F (3, 2), H (4, 5), G (1, 2). The second triangle ha

KIM [24]

Answer:

Correct. He transformed the triangle according to the rule (x, y) → (–y, x)

Step-by-step explanation:

<u>Given</u>

F(3, 2), H(4, 5), G 1, 2)

F'(-2, 3), H'(-5, 4), G'(-2, 1)

<u>Find</u>

Whether the triangle was correctly transformed using the rule ...

(x, y) → (–y, x).

<u>Solution</u>

For point F, we have ...

x = 3

y = 2

Then the rule tells us the point F' should be ...

F' = (-y, x) = (-2, 3)

This is the point Quinton found for F', so his transformation was correct. (We can similarly verify that H' and G' are also correct.)

4 0

2 years ago

The density of air is approximately 1.225 kg/m^3 what is the volume of a soccer ball if the mass of the air it contains is 0.08

svetoff [14.1K]

Using the the density(D)-mass(m)-volume(V) formula $m=DV$ , we can calculate the volume $V= \frac{m}{D}= \frac{0.08}{1.225}= 0.0653 m^{3}$

8 0

2 years ago