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3 years ago

Please help asap!! I will mark you as brainliest :)

Mathematics

1 answer:

zmey [24]3 years ago

3 0

Answer:

the answer is 2.4 million

Step-by-step explanation:

In order to find the rate of change, you first must find the amount that the population grew in the time period:

314-302 = 12

Now you need to calculate the number of years:

2012-2007 = 5

Lastly, divide the amount of change by the number of years:

12/5 = 2.4

The population grew by 2.4 million people per year. Hope this helps! :)

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Maria bikes 15 miles per hour and starts at 10 miles. Glen bikes 20 miles per hour and starts at 0 mile. At what time and distan

Ilya [14]

10/(20-15)=2 hours
20*2=40 miles
15*2+10=40 miles
Glen catch up with Maria after 2 hours at 40 miles

6 0

2 years ago

Write the value of the expression 3^3/3^6

scZoUnD [109]

Answer:

When dividing two values with the same base, we simply subtract the first exponent by the second and keep the original base. 3^3-6=3^-3. This equals out to be 1/27.

Step-by-step explanation:

6 0

3 years ago

Please help me i'm begging you

victus00 [196]

5/6(3/2) = 15/12
5/4
1 and 1/4 hours

5 0

2 years ago

Read 2 more answers

Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim

tino4ka555 [31]

Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{10959}{25} =438.36$

Sum of squares of differences = 175413.76

$S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49$

Population mean, μ = 425 mg/L

Sample mean, $\bar{x}$ = 438.36

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

$H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}$

We use one-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{438.36 - 425}{\frac{85.49}{\sqrt{25}} } = 0.7813$

Now, $t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108$

Since,

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

6 0

3 years ago

Worth 10pts if someone can help!

charle [14.2K]

Answer:

y=-x

I found the points then simplified it down

7 0

2 years ago

Read 2 more answers