Ask question

Ask question

All categories

3 years ago

14

A medical researcher designs a drug to treat high blood pressure. She believes that people of different genders may respond diff

erently to the new drug. What type of sample should be utilized to control for the possibility of different reactions based on gender?
A. stratified random sample
B. systematic sample
C. self-selecting sample
D. random sample

1 answer:

ELEN [110]3 years ago

5 0

Answer:

hello!

Step-by-step explanation:

I say u can go for option B

hope it helps!

You might be interested in

Please please help me on this one!

gizmo_the_mogwai [7]

Answer:

3422 x232

Step-by-step explanation:

3 0

2 years ago

When the members of a family discussed where their annual reunion should take place, they found that out of all the family mem

Doss [256]

Answer:

The total number of family members is 21.

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of these sets.

I am going to say that:

-The set A represents those that would not go to a park.

-The set B represents those who would not go to a beach.

-The set C represents those who would not go to the family cottage.

The value d represents those who would go to all three places.

We have that:

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

In which a are those that would only not go to a park, $A \cap B$ are those who would not got to a park or to the beach, $A \cap C$ are those who would not go to a park or to the famili cottage. And $A \cap B \cap C$ are those that would not go to any of these places.

By the same logic, we have:

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

This diagram has the following values:

$a,b,c,d,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)$

The total number of family members is the sum of all these values:

$T = a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

We start finding the values from the intersection of the three sets

5 would not go to a park or a beach or to the family cottage.

This means that $A \cap B \cap C = 5$

1 would go to all three places. This means that $d = 1$ .

8 would go to neither a park nor the family cottage

This means that:

$A \cap C + (A \cap B \cap C) = 8$

$A \cap C = 3$

8 would go to neither a beach nor the family cottage

$B \cap C + (A \cap B \cap C) = 8$

$B \cap C = 3$

7 would go to neither a park nor a beach

$A \cap B + (A \cap B \cap C) = 7$

$A \cap B = 2$

15 would not go to the family cottage

$C = 15$

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

$15 = c + 3 + 3 + 5$

$c = 4$

12 would not go to a beach

$B = 12$

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$12 = b + 3 + 2 + 5$

$b = 2$

11 would not go to a park

$A = 11$

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

$11 = a + 2 + 3 + 5$

$a = 1$

Now, we can find the total number of family members.

$T = a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

$T = 1 + 2 + 4 + 1 + 2 + 3 + 3 + 5$

$T = 21$

The total number of family members is 21.

5 0

2 years ago

The board of directors of a company knows that the probability that carbon emissions from the company’s factory exceed the permi

jekas [21]

the real answer is D. 0.9132

3 0

2 years ago

Read 2 more answers

If it took Ben 25 minutes to get to work and 55 minutes to get home, how long did he spend traveling to and from Work

Lynna [10]

Answer:

80 minutes... or an hour and 20 minutes

Step-by-step explanation:

55+25 = 80. an hour is 60 minutes long

7 0

3 years ago

Read 2 more answers

Now there is a square city of unknown size with a gate at the center of each side. There is a tree 20 b from the north gate. Tha

Brums [2.3K]

Answer:

The length of each side of the city is 250b

Step-by-step explanation:

Given

$a = 20$ --- tree distance from north gate

$b =14$ --- movement from south gate

$c = 1775$ --- movement in west direction from (b)

See attachment for illustration

Required

Find x

To do this, we have:

$\triangle ADE \sim \triangle ACB$ --- similar triangles

So, we have the following equivalent ratios

$AE:DE = AB:CB$

Where:

$AE = 20\\ DE = x/2 \\ AB = 20 + x + 14 \\ CB = 1775$

Substitute these in the above equation

$20:x/2 = 20 + x + 14: 1775$

$20:x/2 = x + 34: 1775$

Express as fraction

$\frac{20}{x/2} = \frac{x + 34}{1775}$

$\frac{40}{x} = \frac{x + 34}{1775}$

Cross multiply

$x *(x + 34) = 1775 * 40$

Open bracket

$x^2 + 34x = 71000$

Rewrite as:

$x^2 + 34x - 71000 = 0$

Expand

$x^2 + 284x -250x - 71000 = 0$

Factorize

$x(x + 284) -250(x + 284)= 0$

Factor out x + 284

$(x - 250)(x + 284)= 0$

Split

$x - 250 = 0 \ or\ x + 284= 0$

Solve for x

$x = 250 \ or\ x =- 284$

x can't be negative;

So:

$x = 250$

6 0

3 years ago