The volume of a one with a slant height of 50 inches and a diameter of 60 inches (approximate pi

There are 140 apples for every 91 cups of water how many apples are needed for every 1 cup of water

podryga [215]

Answer:

The number of apple for 1 cup of water is 1.5 apples

Step-by-step explanation:

Given as :

The number of cups of water = 91 cups

The number of apples for every 91 cups of water = 140

Let The number of apple for 1 cup of water = n

<u>Now, From unitary method :</u>

∵ For every 91 cups of water , the number of apple needed = 140

∴ For every 1 cup of water , the number of apple needed = $\frac{140}{91}$

I.e n = $\frac{140}{91}$

or, n = 1.5 apples

The number of apple for 1 cup of water = n = 1.5 apples

Hence The number of apple for 1 cup of water is 1.5 apples Answer

5 0

3 years ago

The distance between P and T on the coordinate grid is ___ units. (Input whole numbers only.) Image of a coordinate grid with po

Vlada [557]

That would be 25 units. You can draw it out and you would get 25 units, or you could see the difference in -10 and 15, their distance would be 25 units.

That made no sense.

3 0

4 years ago

Read 2 more answers

Seventy-eight is 26% of what number?

tensa zangetsu [6.8K]

78 / 0.26 = 300 so the answer is 300

8 0

3 years ago

Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the

Alexxandr [17]

Answer:

Step-by-step explanation:

The table can be computed as:

Advertising Expenses ($ million) Number of companies

25 up to 35 4

35 up to 45 19

45 up to 55 27

55 up to 65 16

65 up to 75 9

TOTAL 75

Let's find the probabilities first:

$P(25 - 35) = P \Big(\dfrac{25-50.93}{10.80}$

For 35 up to 45

$P(35 - 45) = P \Big(\dfrac{35-50.93}{10.80}$

For 45 up to 55

$P(45 - 55) = P \Big(\dfrac{45-50.93}{10.80}$

For 55 up to 65

$P(55 - 65) = P \Big(\dfrac{55-50.93}{10.80}$

For 65 up to 75

$P(65 - 75) = P \Big(\dfrac{65-50.93}{10.80}$

Chi-Square Table can be computed as follows:

Expense No of Probabilities(P) Expe $(O-E)^2$ $\dfrac{(O-E)^2}{E}$

compa cted E (n*p)

nies (O)

25-35 4 0.0612 75*0.0612 = 4.59 0.3481 0.0758

35-45 19 0.2218 75*0.2218 = 16.635 5.5932 0.3362

45-55 27 0.3568 75*0.3568 = 26.76 0.0576 0.021

55-65 16 0.2552 75*0.2552 = 19.14 9.8596 0.5151

65-75 9 0.0839 75*0.0839 = 6.2925 7.331 1.1650

$\sum \dfrac{(O-E)^2}{E}= 2.0492$

Using the Chi-square formula:

$X^2 = \dfrac{(O-E)^2}{E} \\ \\ Chi-square \ X^2 = 2.0942$

Null hypothesis:

$H_o: \text{The population of advertising expenses follows a normal distribution}$

Alternative hypothesis:

$H_a: \text{The population of advertising expenses does not follows a normal distribution}$

Assume that:

$\alpha = 0.02$

degree of freedom:

= n-1

= 5 -1

= 4

Critical value from $X^2 = 11.667$

Decision rule: To reject $H_o \ if \ X^2$ test statistics is greater than $X^2$ tabulated.

Conclusion: Since $X^2 = 2.0942$ is less than critical value 11.667. Then we fail to reject $H_o$

6 0

3 years ago

3(x - 5) – 3 = 5(x - 2) + 2x

Arlecino [84]

Answer:

x=−2

Step-by-step explanation: Step 1: Simplify both sides of the equation.

3(x−5)−3=5(x−2)+2x

(3)(x)+(3)(−5)+−3=(5)(x)+(5)(−2)+2x(Distribute)

3x+−15+−3=5x+−10+2x

(3x)+(−15+−3)=(5x+2x)+(−10)(Combine Like Terms)

3x+−18=7x+−10

3x−18=7x−10

Step 2: Subtract 7x from both sides.

3x−18−7x=7x−10−7x

−4x−18=−10

Step 3: Add 18 to both sides.

−4x−18+18=−10+18

−4x=8

Step 4: Divide both sides by -4.

−4x

−4

=

8

−4

x=−2

6 0

3 years ago