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2 years ago

5

A body builder consumed 4,400 calories yesterday. Today, he consumed 3,600 calories . What is the percent decrease in number of

calories consumed? Round to the nearest percent

1 answer:

IgorC [24]2 years ago

5 0

Answer:

it's about 18 percent decrease**** so 20 percent to round up

Step-by-step explanation:

You might be interested in

Find the diameter of the circle with the given circumference use 3.14 C=18

algol [13]

The formula for the circumference of a circle is pi * r * 2

First we solve for r (the radius)
pi*r*2 = C
pi*r*2 = 18

Divide 18 by 2
pi*r = 9

Then divide 9 by 3.14
pi= 9/3.14 which is 2.866

The diameter is 2 times length of the radius so multiply 2.866 by 2
The diameter is about 5.732
(And you can round of course)

Hope that helped

3 0

2 years ago

In 1960, the population of a town was 14 thousand people. Over the course of the next 50 years, the town grew at a rate of 30 pe

andre [41]

SOLUTIONS

Given: Assume y is the population

$\begin{gathered} t=0=1960,y=14030 \\ t=1=1961,y=14060 \\ so\text{ b = 14000} \\ so\text{ k = 30} \end{gathered}$

The town grow at the rate of 30 people per year.

$y=kt+14000$ $y=30t+14000$

(A) The population predicted to be in 2030 will be

$\begin{gathered} 2030-1960=70 \\ y=30k+14000 \\ y=30(70)+14000 \\ y=2100+14000 \\ y=16100people \end{gathered}$

(B) so when y = 15000, find t

$\begin{gathered} y=15000 \\ y=30t+14000 \\ 15000=30t+14000 \\ 15000-14000=30t \\ 1000=30t \\ t=\frac{1000}{30} \\ t\approx33 \\ t=1960+33 \\ t=1993 \end{gathered}$

3 0

4 months ago

US average math SAT scores follow a normal distribution with a mean of 505 and a standard deviation of 112. A sample of 64 enter

Hitman42 [59]

Answer:

The claim that the scores of UT students are less than the US average is wrong

Step-by-step explanation:

Given : Sample size = 64

Standard deviation = 112

Mean = 505

Average score = 477

To Find : Test the claim that the scores of UT students are less than the US average at the 0.05 level of significance.

Solution:

Sample size = 64

n > 30

So we will use z test

$H_0:\mu \geq 477\\H_a:\mu < 477$

Formula : $z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

$z=\frac{505-477}{\frac{112}{\sqrt{64}}}$

$z=2$

Refer the z table for p value

p value = 0.9772

α=0.05

p value > α

So, we accept the null hypothesis

Hence The claim that the scores of UT students are less than the US average is wrong

3 0

2 years ago

8.4×0.5 showing work

Katen [24]

Can u add a picture?

8 0

2 years ago

Gala2k [10]

Answer:

The second one.

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers