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BlackZzzverrR [31]

3 years ago

15

What is the multiplicative rate of change for the exponential function graphed to the left

2 answers:

Ksenya-84 [330]3 years ago

7 0

Answer:

3

Step-by-step explanation:

kifflom [539]3 years ago

6 0

Looking at the given points on the right side from (0,2) to (1,6) for 1 increase in X ( 1-0=1) the Y value increases by 3 ( 6/2 = 3)

This same increase happens for th other two points: 18/6 = 3

54 / 18 = 3

The rate of increase is 3.

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Can someone help me with this please

ryzh [129]

Answer:

y=4

Step-by-step explanation:

You can ignore the 3 and 6 because this has no y value change(rise over run)

This is a straight horizontal line so the equation is y=4.

Slope=0

4 0

2 years ago

14 POINTS 7x - 3 = 18

Svetlanka [38]

Subtract 3 first so it would be 7x=15
then divide by 7 so x would be approximately 2.14

5 0

3 years ago

Read 2 more answers

pipe A is a inlet pipe filling the tank at 8000l/hr and pipe B is the outlet pipe which empties the tank in 3 hours.the capacity

lawyer [7]

Well , in 3 hours , the amount of the tank that will be filled with pipe A = 8,000 x 3 = 24,000 L

Pipe could empty that 24,000 L in 3 hours.

So the capacity is around 6 - 8 L . .. . we need additional information to find out for sure

hope this helps

8 0

3 years ago

1.) Consider the right triangle below in which a = 5 and b = 5.

patriot [66]

Answer:

c=7.07

Step-by-step explanation:

c²=a²+b²

c²=5²+5²

c²=25+25

c²=50

c=√50

c=7.07

B. c=7.07 (to nearest hundredth)

C. c=√50= 5√2 in radical form

3 0

2 years ago

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