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3 years ago

A turtle traveled 2/5 of a mile in one hour. how many miles would it travel in 6 hours?

Mathematics

2 answers:

Delvig [45]3 years ago

6 0

2/5 plus 2/5....
2 2/5 is your answer

fomenos3 years ago

5 0

3 miles... Brainiest Please!!!

You might be interested in

(57x2:3)+32x450-[39-(4+2):2[+8=

tigry1 [53]

Answer:

57×2/3+32×450-[39-4+2]/2+8

57×2/3+32×450-(39-6/2)+8

57×2/3+32×450-36+8

57×2/3+14400-28

57×2/3+14372

38+14372

14410

8 0

2 years ago

The initial cost to drill a well is $210000. the daily rate to operate the well is $500. The current selling price is $65 per ba

SVETLANKA909090 [29]

Hi there!

There is an initial cost of $210,000 (which you're just going to pay once)

Then it costs $500 per day to operate.

The number of days is represented by "x".

Your equation in function notation should look like this :

f(x) = 500x + 210,000

There you go! I really hope this helped, if there's anything just let me know! :)

8 0

3 years ago

In one paragraph explain step by step how to subtract 7 from 102

Anna11 [10]

To subtract 7 from 102, you have to regroup. You subtract 2 from 100, and 2 from 7, so it is 100-5. 100-5 is 95. The answer is 95. If you don't understand, tell me in the comments. This is just one method of subtracting.

8 0

3 years ago

Read 2 more answers

In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s

Oksi-84 [34.3K]

Answer:

a) $\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

b) From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

c) $P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

$P(Z\geq2.070)=1-P(Z$

Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

$\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4$

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

$\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

Part b

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

Part c

For this case we want this probability:

$P(\bar X \geq 285)$

And we can use the z score defined as:

$z=\frac{\bar x -\mu}{\sigma_{\bar x}}$

And using this we got:

$P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

And using a calculator, excel or the normal standard table we have that:

$P(Z\geq2.070)=1-P(Z$

8 0

3 years ago

Kelly deposited $4,205 in an account for 3.5 years. She earned $662.29 in interest. What interest rate did the bank pay her?

mylen [45]

Answer:

r = 4.5%

Step-by-step explanation:

Formula:

I = Prt

r = I/(Pt)

Given:

I = 662.29

P = 4205

t = 3.5

Work:

r = I/(Pt)

r = 662.29/(4205 * 3.5)

r = 662.29/14717.5

r = 0.045

r = 4.5%

5 0

2 years ago