Please help asap!!!!

Homemade applesauce is one of Jared's favorite foods, so he stocked up on 10 pounds of

Vanyuwa [196]

Answer:

<h2>5 quarts</h2>

Step-by-step explanation:

Let's think through this bit by bit.

10 lbs * 16 oz/lb = 160 oz. of apples.

In the proportion below, we can see we simply divide by 8, in terms of magnitude.

8 oz -> 1 cup

160 oz -> 160/8 -> 20 cup

20 cups! Is it the answer? NO! They're asking for it in units of quarts, not cups.

We must remember that 1 quart = 4 cups.

20 cups * 1/4 qt/cups = 5 quarts

8 0

3 years ago

How many ways can a cook arrange 9 of her 12 spice jars on a spice rack?

pshichka [43]

It depends if the 12 spice jars are all different.

If they are:

_ _ _ _ _ _ _ _ _ (9 spaces)

12*11*10*9*8*7*6*5*4

12 possibilities for first one, 11 for second, etc...

Multiply out to get:
79833600

5 0

3 years ago

Read 2 more answers

What is the sum or difference?

dlinn [17]

1. Ans: Option (B) $-6x^4$

Explanation:
Given: $2x^4 - 8x^4$
Take the common out:
=> $x^4(2 - 8)$

Hence: => $[tex]-6x^4$ [/tex] (Option B)

2. Ans: Option (D) $-3y^5$

Explanation:
Given: $6y^5 - 9y^5$
Take the common term(s) out:
=> $y^5(6-9)$

Hence: => $-3y^5$ (Option D)

3. Ans: Option (B) $9x^2 - 12x + 6$ quadratic trinomial

Explanation:
Given: $6 - 12x + 13x^2 - 4x^2$

The standard form of polynomial function must have the highest powered value at the start, then the second highest and so on.

=> $13x^2 - 4x^2 - 12x + 6$
=> $9x^2 - 12x + 6$ (Option B)

4. Ans: Option (C) $7.3x^2 + 0.8x + 1.3$

Explanation:
In order to find the total number of Common and Endler's guppies, you need to add both Common and Endler's guppies' polynomials, as follows:

Common guppies: $3.1x^2 +6.0x + 0.3$ 
Endler's guppies: $4.2x^2 - 5.2x + 1.0$
(add both)
-----------------------------------------------------------------
Total number: $7.3x^2 +0.8x + 1.3$
-----------------------------------------------------------------
Hence the ans is Option(C) $7.3x^2 + 0.8x + 1.3$

5. Ans: Option (D) $2x^2(133 - 18 \pi )$

Explanation:
First let's find the total area of the yard:
Total Area of the Yard = 14x * 19x = $266x^2$

Now the area of the circular fountain:
Area of the Circular Fountain = $\pi r^2$
Since, r=6x
Therefore,
Area of the Circular Fountain = $\pi (6x)^2 = 36 \pi x^2$

Now the final Area of the yard would be:
Final area of the Yard = Total Area of the Yard - Area of the Circular Fountain
Final area of the Yard = $266x^2$ - $36 \pi x^2$
=> Final area of the Yard = $2x^2(133 - 18 \pi)$ (Option D)

6. Ans: Option (A) $56x^2$

Explanation:
First let's find the total area of the lot:
Total Area of the lot= 6x * 10x = $60x^2$

Now the area of the stadium:
Area of the stadium = 1x * 4x = $4x^2$

Now the final Area of the lot would be:
Final area of the lot= Total Area of the lot - Area of the stadium
Final area of the lot= $60x^2$ - $4x^2$
=> Final area of the lot = $56x^2$ (Option A)

-i

4 0

3 years ago

Read 2 more answers

Write as a product of two polynomials. 2(3-b)+5(b-3)^2

viktelen [127]

Answer:

$5b^{2}-32b+51$

Step-by-step explanation:

$2(3-b)+5(b-3)^{2}$

$=6-2b+5b^{2}-30b+45$

$=5b^{2}-32b+51$

4 0

2 years ago

A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri

nlexa [21]

Answer:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday Tuesday Wednesday Thursday Friday Saturday Total

12 9 11 10 9 9 60

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{60}{6}= 10$ and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

$\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8$

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

4 0

3 years ago