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3 years ago

13

-3 4/7 divide -1 1/14

1 answer:

Sedaia [141]3 years ago

8 0

Answer:

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Step-by-step explanation:

You might be interested in

The cost of 3 ibs of turkey is $9.81. What is the cost per pound for the turkey?

butalik [34]

3 = $9.81

1 = $?

To find this you have to do 9.81/3.

Hope this helps and have a nice day!!

3 0

3 years ago

Find the area of the figure shown. Use 3.14 to approximate pi

Nina [5.8K]

Alrighty

we gots 1/4 circle and paralellogram

area circle=pi times radius^2
area paralelogram=base times height

1/4 circle is
radius is 10 so
area part circle is 1/4 times pi times 10^2=100/4pi=25pi=78.5 m²

area paralllogram is
10 times 3 because 3 is height and 10 is base length or 30

30+78.5=108.5 m² is the area

4 0

3 years ago

-2x-7≥41 in algaebra for

dexar [7]

Answer:

They are equal. x= -24

Step-by-step explanation:

-2x - 7 ≥ 41

First you isolate the variable

-2x -7 ≥ 41

+7 +7

-2x≥48

then divide each side by -2

x≥-24

Then go back and plug in

-2(-24)-7=

48-7=41

Therefore

41=41

4 0

3 years ago

pin Co. has $52,000 in its Cash account, $20,000 in its Inventory account, and $12,000 in its Notes Payable (short-term) account

KatRina [158]

Answer:

60000

Step-by-step explanation:

Assets - liabilities = share holder equity

(52,000+20,000)−12,000=60000

5 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