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2 years ago

Solve 2a 2 - 3a - 2 = 0 {-1/2, 2} {1/2, -2} {}

2 answers:

8 0

Apply slip and slide
a^2-3a-4
(a-4)(a+1)
(a-2)(2a+1)

Find zeros
a-2=0
a=2

2a+1=0
2a=-1
a=-1/2

Final answer: {-1/2, 2}

Igoryamba2 years ago

8 0

The answer is {-1/2, 2}

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Evaluate. Write your answer as an integer or as a decimal rounded to the nearest hundredth. tan 26° = ___

barxatty [35]

Answer:

0.49

Explanation:

Given the expression:

tan 26°

Evaluating using calculator

tan 26° = 0.4877

Hence the decimal rounded to the nearest hundredth is 0.49

4 0

6 months ago

Read 2 more answers

a survey amony freshman at a certain university revealed that the number of hours spent studying the week before final exams was

Marat540 [252]

Answer:

Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

Step-by-step explanation:

We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.

A sample of 36 students was selected.

Let $\bar X$ = sample average time spent studying

The z-score probability distribution for sample mean is given by;

Z = $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ ~ N(0,1)

where, $\mu$ = population mean hours spent studying = 25 hours

$\sigma$ = standard deviation = 15 hours

n = sample of students = 36

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < $\bar X$ < 30 hours)

P(29 hours < $\bar X$ < 30 hours) = P( $\bar X$ < 30 hours) - P( $\bar X$ $\leq$ 29 hours)

P( $\bar X$ < 30 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ < $\frac{ 30-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z < 2) = 0.97725

P( $\bar X$ $\leq$ 29 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ $\leq$ $\frac{ 29-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z $\leq$ 1.60) = 0.94520

So, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.

Therefore, P(29 hours < $\bar X$ < 30 hours) = 0.97725 - 0.94520 = 0.0321

Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

7 0

2 years ago

What one half subtracted by one fourth

WINSTONCH [101]

One half subtracted by one fourth is one fourth.
1/2 - 1/4 = 1/4

4 0

2 years ago

Read 2 more answers

A class at Middlebury school collected date on the types of movies students prefer. Complete each statement using the table.

likoan [24]

Based on the information of the table, you have:

1. The ratio is 105/150. By simplifying you get for the ratio 7/10.

2. The students that prefer action movies are 75+90 = 165 and the total numbe of students is 180+240 = 420. Then, the fraction of students who prefer action movies is:

165/420 = 11/28

3. The fraction of seventh graders students that prefer action movies is:

75/180 = 5/12

4. The percent of student that prefer comedy is:

105 + 150 = 255 total student that prefer comedy

420 total number of students

the fraction is:

(x/100)420 = 255

solve for x:

x = 255(100/420)

x = 60.71

the percent of students is 60.71%

5. The percent of eighth graders student who prefer action moveis is:

(x/100)240 = 90

x = 90(100/240)

x = 37.5

the percent of students is 37.5%

6. To determine which from the given grades has the greatest percent of student that prefer action movies, calculate the percent of student in seventh-grade:

(x/100)180 = 75

x = 75(100/180)

x = 41.66

the percent of student is 41.66%

then, seventh grade has the greatest percent of student that prefer action movies.

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1 month ago

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mamaluj [8]

Answer: .0142%

Step-by-step explanation:

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