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zhannawk [14.2K]
2 years ago
13

A recipe for whole wheat pizza crust calls for 1 1/2 cups of all purpose flour. Jeremy has 5 7/4 cups of all purpose flour. If h

e makes the recipe, how much flour will he have left?
​
Mathematics
1 answer:
dybincka [34]2 years ago
6 0

Answer:

Jeremy has 5 1/4 cups of all purpose flour after preparing his recipe.

Step-by-step explanation:

Since the recipe for whole wheat pizza crust calls for 1 1/2 cups of all purpose flour, 1.5 cups of that product is required to create the meal in question.

Now, since Jeremy has 5 7/4 of all purpose flour, it can be stated that he has 6.75 cups of it (7/4 = 1.75).

Therefore, to determine the amount of flour that Jeremy will have left over after preparing the recipe, it is necessary to perform the following calculation:

6.75 - 1.5 = X

5.25 = X

So since 0.25 is equal to 1/4, Jeremy has 5 1/4 cups of all purpose flour after preparing his recipe.

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The slope of the line that passes through the points (-2, -3) and (8, -3) is....
andre [41]

Answer:

<h2>Zero</h2>

Step-by-step explanation:

\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}\\\\\left(x_1,\:y_1\right)=\left(-2,\:-3\right),\:\\\left(x_2,\:y_2\right)=\left(8,\:-3\right)\\\\m=\frac{-3-\left(-3\right)}{8-\left(-2\right)}\\\\m=0

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3 years ago
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Health insurance benefits vary by the size of the company (the Henry J. Kaiser Family Foundation website, June 23, 2016). The sa
xxMikexx [17]

Answer:

\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221

Now we can calculate the degrees of freedom for the statistic given by:

df=(rows-1)(cols-1)=(3-1)(2-1)=2

And we can calculate the p value given by:

p_v = P(\chi^2_{2} >19.221)=0.000067

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example \alpha=0.05, we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent 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

Assume the following dataset:

Size Company/ Heal. Ins.   Yes   No  Total

Small                                      32   18    50

Medium                                 68     7    75

Large                                     89    11    100

_____________________________________

Total                                     189    36   225

We need to conduct a chi square test in order to check the following hypothesis:

H0: independence between heath insurance coverage and size of the company

H1:  NO independence between heath insurance coverage and size of the company

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{total col * total row}{grand total}

And the calculations are given by:

E_{1} =\frac{50*189}{225}=42

E_{2} =\frac{50*36}{225}=8

E_{3} =\frac{75*189}{225}=63

E_{4} =\frac{75*36}{225}=12

E_{5} =\frac{100*189}{225}=84

E_{6} =\frac{100*36}{225}=16

And the expected values are given by:

Size Company/ Heal. Ins.   Yes   No  Total

Small                                      42    8    50

Medium                                 63     12    75

Large                                     84    16    100

_____________________________________

Total                                     189    36   225

And now we can calculate the statistic:

\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221

Now we can calculate the degrees of freedom for the statistic given by:

df=(rows-1)(cols-1)=(3-1)(2-1)=2

And we can calculate the p value given by:

p_v = P(\chi^2_{2} >19.221)=0.000067

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example \alpha=0.05, we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

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never [62]

Isolate the x in both cases. What you do to one side, you do to the other.


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Subtract 4 from both sides


11x + 4 (-4) < 15 (-4)


11x < 11


isolate the x, divide 11 from both sides


11x/11 < 11/11


x< 1


x < 1 is your answer for the first one

-------------------------------------------------------------------------------------------------------------------


Again, isolate the x.


12x - 7 > -25


Add 7 to both sides


12x - 7 (+7) > -25 (+7)


12x > -18


Isolate the x, divide 12 from both sides


12x/12 > -18/12


x > -1.5 is your answer for the second one.


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hope this helps

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