(04.04 MC)

I NEED HELP ON THE ONES THAT AREN'T CROSSED OFF!!!!!!!!!!!!!!!!!!!!! PLEASEEEEEEEEEEEEEEEEEEEEE ( i already did 8)

jonny [76]

The computation of the word problem shows that the oranges will John have in 12 weeks will be 70 oranges.

<h3>How to solve the word problem?</h3>

The word problem will be that John has 10 oranges and buys 5 more oranges every week. How many oranges will John have in 12 weeks.

The computation will be:

= 10 + 5x

where x= number of weeks.

= 10 + 5x

= 10 + 5(12)

= 10 + 60

= 70 oranges

Learn more about word problems on:

brainly.com/question/13818690

#SPJ1

6 0

3 years ago

The College Board SAT college entrance exam consists of three parts: math, writing and critical reading (The World Almanac 2012)

Wittaler [7]

Answer:

Yes, there is a difference between the population mean for the math scores and the population mean for the writing scores.

Test Statistics = $\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$ .

Step-by-step explanation:

We are provided with the sample data showing the math and writing scores for a sample of twelve students who took the SAT ;

Let A = Math Scores ,B = Writing Scores and D = difference between both

So, $\mu_A$ = Population mean for the math scores

$\mu_B$ = Population mean for the writing scores

Let $\mu_D$ = Difference between the population mean for the math scores and the population mean for the writing scores.

Null Hypothesis, $H_0$ : $\mu_A = \mu_B$ or $\mu_D$ = 0

Alternate Hypothesis, $H_1$ : $\mu_A \neq \mu_B$ or $\mu_D \neq$ 0

Hence, Test Statistics used here will be;

$\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$ where, Dbar = Bbar - Abar

$s_D$ = $\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}$

n = 12

Student Math scores (A) Writing scores (B) D = B - A

1 540 474 -66

2 432 380 -52

3 528 463 -65

4 574 612 38

5 448 420 -28

6 502 526 24

7 480 430 -50

8 499 459 -40

9 610 615 5

10 572 541 -31

11 390 335 -55

12 593 613 20

Now Dbar = Bbar - Abar = 489 - 514 = -25

Bbar = $\frac{\sum B_i}{n}$ = $\frac{474+380+463+612+420+526+430+459+615+541+335+613}{12}$ = 489

Abar = $\frac{\sum A_i}{n}$ = $\frac{540+432+528+574+448+502+480+499+610+572+390+593}{12}$ = 514

∑ $D_i^{2}$ = 22600 and $s_D$ = $\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}$ = $\sqrt{\frac{22600 - 12*(-25)^{2} }{12-1} }$ = 37.05

So, Test statistics = $\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$

= $\frac{-25 - 0}{\frac{37.05}{\sqrt{12} } }$ follows $t_1_1$ = -2.34

Now at 5% level of significance our t table is giving critical values of -2.201 and 2.201 for two tail test. Since our test statistics doesn't fall between these two values as it is less than -2.201 so we have sufficient evidence to reject null hypothesis as our test statistics fall in the rejection region .

Therefore, we conclude that there is a difference between the population mean for the math scores and the population mean for the writing scores.

8 0

3 years ago

Solve for the variable 4/2 = 9/x A x = 2/9 b x=4 C =4.5 D x= 18

Lesechka [4]

Answer:

$\frac{4}{2} = \frac{9}{x} \\$

perform cross multiplication:

$(4 \times x) = 9 \times 2 \\ 4x = 18 \\ → \: { \tt{x = 4.5}}$

5 0

3 years ago

Một ủy ban an toàn kiểm tra ngẫu nhiên 900 công nhân xây dựng trong thời gian làm việc, thấy có 48 công nhân không mang mũ bảo h

Usimov [2.4K]

Answer:

The percentage of workers not wearing the helmets is 5.3 %.

Step-by-step explanation:

A safety committee randomly examined 900 construction workers during their work, and found that 48 workers were not wearing helmets. Estimate the percentage of workers who do not wear protective masks during their working time with 98% confidence

total workers = 900

Not wearing helmet = 48

Percentage which are not wearing the helmets

= $\frac{48}{900}\times 100 = 5.3 %$ %

6 0

3 years ago

Factor completely: −3x2 + 6x − 9

denis23 [38]

Factor out the -3
(-3)(x^2-2x+3)

8 0

3 years ago

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