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3 years ago

Pleasee help with these questions

Mathematics

1 answer:

maw [93]3 years ago

6 0

Answer:

6. 3.14 7. 172.03 8. 78.54 9. 380.13 10. 19.63 11.28.27

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PLEASE HELP VERY IMPORTANT I REALLY NEEDS HELP

strojnjashka [21]

[see picture link]Area of part 1:A1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2Area of part 2:A2=length*width=(8+2)*1=10 ft2Area of part 3:A3=length*width=8*2=16 ft2
Picture: https://us-static.z-dn.net/files/d30/9033549987fa60cf57fe05f71424f322.gif

6 0

3 years ago

9.6x10^85/3x10^63 in scientific notation

Novosadov [1.4K]

Answer:

Step-by-step explanation:

Given the expression 9.6x10^85/3x10^63, we are to write it on scientific notation as shown:

9.6x10^85/3x10^63

= (9.8/3) * (10^85/10^63)

= (9.8/3) * 10^{85-63}

= (9.8/3) *10^22

= 3.27 *10^22

<em>Hence the expression in scientific notation is 3.27*10^22</em>

6 0

3 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

What is the mean of this data set? 12,17,16,10,20,13,14,14,12,12,19,18

Gemiola [76]

Answer:

14.75 :) please give me brainliest :)

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What is the inverse of the function f(x) = 2x – 10?

noname [10]

Answer:

The inverse f-1(x) = ( x + 10)/2.

Step-by-step explanation:

f(x) = 2x - 10

Find x in terms of f(x):

2x = f(x) + 10

x = (f(x) + 10))/ 2

Replace x by the inverse f-1(x) and f(x) by x:

f-1(x) = (x + 10)/2.

7 0

3 years ago