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3 years ago

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Lucy correctly answered 70% of the questions on her science homework.she counted 14 correct answers on her paper .if each questi

on is of a equal value ,how many questions did lucy have on her science homework

1 answer:

disa [49]3 years ago

4 0

Lucy has 20 questions in her science home work

<em><u>Solution:</u></em>

Given that,

Lucy correctly answered 70% of the questions on her science homework

She counted 14 correct answers on her paper

<em><u>To find: Number of questions in her science home work</u></em>

Let "x" be the number of questions in her science home work

From given we can say,

Lucy correctly answered 70% of questions

She counted 14 correct answers

70 % of total questions = 14 correct answers

70 % of x = 14

$70 \% \times x = 14\\\\\frac{70}{100} \times x = 14\\\\0.70x = 14\\\\x = \frac{14}{0.70}\\\\x = 20$

Thus Lucy has 20 questions in her science home work

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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.

<em> Null Hypothesis, </em> $H_0$ <em> : </em> $\mu_A = \mu_B$ <em> or </em> $\mu_D$ <em> = 0 </em>

<em> Alternate Hypothesis, </em> $H_1$ <em> : </em> $\mu_A \neq \mu_B$ <em> or </em> $\mu_D \neq$ <em> 0</em>

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

<em>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 .</em>

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

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3 years ago

You invest $500 in an account that has a annual interest rate of 5%, compounded quarterly for four years. What is the equivalent

posledela

Answer:

1.25% and 16 times

Step-by-step explanation:

Since the interest is compounded quarterly it will be compounded 4 times a year. So 4 x 4 is 16, so it will be compounded 16 times.

Then you have to divide the 5% by four to get how much will be compounded each quarter. So, (0.05 / 4) = 0.0125, which is 1.25%.

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Answer:

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You can assign 4 numbers (0, 1, 2, 3) to the event "there is a major storm"

in this way, you have 40% of having a storm.

and the other 6 numbers (4, 5, 6, 7, 8, 9) to the event "there is not a major storm". This means that we have 60% of not having a storm.

now, when you roll the dice you can see if a year there will be a storm or not.

Now, you roll the dice 5 times (for the 5 years) and take note of the results and the number of storms in those 5 rolls.

Now do the same thing a bunch of times, at least 25 times.

Now, you recorded the results in each set of rolls, now see the number of sets that have at least years with storms.

Take that number and divide it by the total number of sets of data (in this case 25, for example)

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Find the GCF (Greatest Common Factor)

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Factor out the GCF ( Write the GCF first. Then, in parentheses, divide each term by the GCF)

3(3x^2/3 + -12x/3 - 15/3)

Simplify each term in parentheses

3(x^2 - 4x - 5)

Factor x^2 - 4x - 5

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3 years ago

Where would 13/4 be on the number line

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3 years ago

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