There are 9 students in a class: 7 boys and 2 girls.

Which answer !!!!!!!!!

ycow [4]

The first answer, answer A. It's easy.

3 0

3 years ago

What the square root

MrRissso [65]

Answer:

-7.07106

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Marshall invest $4500 into an account bearing 6% interest compounded semiannually. How much interest will be added into this acc

Sphinxa [80]

Answer:

$1547.62

Step-by-step explanation:

The principal Marshall invested is $4500.

The rate of interest is 6%

The compound interest formula is

$A=P(1+r)^n$

We substitute P=4500,r=0.06 and t=5 to obtain:

$A=4500(1+ \frac{0.06}{2} )^ {2 \times 5}$

We simplify to get:

$A=4500(1+ 0.03)^ {10} = 4500 \times {1.03}^{10}$

This gives us:

$A=6047.62$

The interest after 5 years is

$6047.62 - 4500 = 1547.62$

4 0

2 years ago

What is the slope of the line ? please help me fast

ValentinkaMS [17]

The slope of the line is 3/2 i believe

4 0

3 years ago

The table shows a proportional relationship between the mass, in kilograms

OlgaM077 [116]

Answer:

C) Dog: 40 kg, Medicine: 1.6 mL

D) Dog: 35 kg, Medicine: 1.4 mL

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $k=\frac{y}{x}$ or $y=kx$

Let

x ---> the mass, in kilograms of a dog

y ---> milliliters of flea medicine a veterinarian prescribes

Find the value of the constant of proportionality k for the given data

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

For x=50 kg, y=2 mL ----> $k=\frac{2}{50}=0.04\ mL/kg$

For x=10 kg, y=0.4 mL ----> $k=\frac{0.4}{10}=0.04\ mL/kg$

For x=5 kg, y=0.2 mL ----> $k=\frac{0.2}{5}=0.04\ mL/kg$

so

The linear direct equation is equal to

$y=0.04x$

Verify each choice

A) Dog: 15 kg, Medicine: 0.9 mL

Divide the milliliters of medicine by the kilograms of the dog and compare the result with the constant of proportionality

$\frac{0.9}{15}= 0.06\ mL/kg$

so

$0.06 \neq 0.04$

therefore

These numbers could not be used as the missing values in the table

B) Dog: 25 kg, Medicine: 0.8 mL

Divide the milliliters of medicine by the kilograms of the dog and compare the result with the constant of proportionality

$\frac{0.8}{25}= 0.032\ mL/kg$

so

$0.032 \neq 0.04$

therefore

These numbers could not be used as the missing values in the table

C) Dog: 40 kg, Medicine: 1.6 mL

Divide the milliliters of medicine by the kilograms of the dog and compare the result with the constant of proportionality

$\frac{1.6}{40}= 0.04\ mL/kg$

so

$0.04 = 0.04$

therefore

These numbers could be used as the missing values in the table

D) Dog: 35 kg, Medicine: 1.4 mL

Divide the milliliters of medicine by the kilograms of the dog and compare the result with the constant of proportionality

$\frac{1.4}{35}= 0.04\ mL/kg$

so

$0.04 \neq 0.04$

therefore

These numbers could be used as the missing values in the table

E) Dog: 20 kg, Medicine: 1.2 mL

Divide the milliliters of medicine by the kilograms of the dog and compare the result with the constant of proportionality

$\frac{1.2}{20}= 0.06\ mL/kg$

so

$0.06 \neq 0.04$

therefore

These numbers could not be used as the missing values in the table

5 0

3 years ago