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andrew-mc [135]

3 years ago

11

If you have 18 students and all are paired up at least once how many projects in the year

2 answers:

devlian [24]3 years ago

5 0

9. A pair is two. 18 divided by two is 9.

KatRina [158]3 years ago

4 0

It would be 9 because you get half the number of projects because 18 students are partners and if you divide that you would have 9 groups of 2 and if you multiply that by the number of projects you would get the number of projects (ex:18 dividend by 2 it's 9 then if there is three projects assigned you multiply 9 by 3 to get 27 projects got it?

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Marta_Voda [28]

Answer:Where is the graph?

Step-by-step explanation:

6 0

3 years ago

The Browns are moving across the country. Mr. Brown leaves 3.5 hours before Mrs. Brown. If he averages 45 mph and she averages 9

Alex Ar [27]

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3 years ago

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Viefleur [7K]

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5 0

3 years ago

The profit, in dollars, of a small business can be modeled by the function P(x)=0.3x^2+7x-40, where c is the number of units sol

ra1l [238]

Answer:

The correct answer is 10 units.

Step-by-step explanation:

Profit function of a small business is given by P(x) = 0.3 $x^{2}$ + 7x - 40, where x is the number of units sold.

The small business intend to make a profit of $60.

To find out the number of units the business has to sell in order to have a profit of $60 is given by,

P(x) = 0.3 $x^{2}$ + 7x - 40 = 60

⇒ 0.3 $x^{2}$ + 7x - 100 = 0

⇒ 3 $x^{2}$ + 70x - 1000 = 0

⇒ x = -70 ± $\sqrt{(70)^{2} + 12000}$ × $\frac{1}{6}$

⇒ x = -70 ± 130 × $\frac{1}{6}$

⇒ x = 10 or - $\frac{200}{6}$

Quantity sold cannot be negative giving us the value of x as 10 units.

6 0

3 years ago

You would like to know whether forwards in a basketball league average the same (or different) numbers of points than the overal

Akimi4 [234]

Answer:

The critical t-value for such a test given an alpha level of 0.05 is 2.26

Step-by-step explanation:

Null hypothesis : $H_0:\mu = 7.5$

Alternate hypothesis : $H_a:\mu \neq 7.5$

Population mean = $\mu = 7.5$

Data : 10, 9, 6, 11, 13, 14, 9, 9, 9, and 10

Mean = $\bar{x}=\frac{\text{Sum of all observations}}{\text{no. of observations}}$

Mean = $\frac{10+9+6+11+13+14+9+9+ 9+ 10}{10}$

Mean =10

Standard deviation : $\sqrt{\frac{\sum(x-\bar{x})^2}{n}}$

Standard deviation : $\sqrt{\frac{(10-10)^2+(9-10)^2+(6-10)^2+(11-10)^2+(13-10)^2+(14-10)^2+(9-10)^2+(9-10)^2+(9-10)^2+(10-10)^2}{10}}$

Standard deviation s :2.144

$t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}t = \frac{10-7.5}{\frac{2.144}{\sqrt{10}}}t=3.687$

Df = n-1 = 10-1 = 9

T critical = $t_{(df,\alpha)}=t_(9,0.05)=2.26$

T calculated > T critical

So, We failed to accept null hypothesis

Hence the critical t-value for such a test given an alpha level of 0.05 is 2.26

5 0

3 years ago