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1 year ago

Simplify the following expression.

Mathematics

2 answers:

kobusy [5.1K]1 year ago

6 0

Step-by-step explanation:

(9x2 - 2x +7) +(4x2+7x +2)

9x2 - 2x +7 +4x2 +7x +2

13x2 +5x +9

SSSSS [86.1K]1 year ago

6 0

Answer:

A. 13x2 + 5x + 9

Step-by-step explanation:

use math-way it helps with problems like this

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Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.

So, 90% confidence interval for the population proportion, p is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

90% confidence interval for p = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

$0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }$

$\sqrt{n} = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}$

$\sqrt{n}$ = 54.79

n = $54.79^{2}$

n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

5 0

1 year ago

During the 2015-16 NBA season, J.J. Redick of the Los Angeles Clippers had a free throw shooting percentage of 0.901 . Assume th

ser-zykov [4K]

Answer: 0.5898

Step-by-step explanation:

Given : J.J. Redick of the Los Angeles Clippers had a free throw shooting percentage of 0.901 .

We assume that,

The probability that .J. Redick makes any given free throw =0.901 (1)

Free throws are independent.

So it is a binomial distribution .

Using binomial probability formula, the probability of getting success in x trials :

$P(X=x)^nC_xp^x(1-p)^{n-x}$

, where n= total trials

p= probability of getting in each trial.

Let x be binomial variable that represents the number of a=makes.

n= 14

p= 0.901 (from (1))

The probability that he makes at least 13 of them will be :-

$P(x\geq13)=P(x=13)+P(x=14)$

$=^{14}C_{13}(0.901)^{13}(1-0.901)^1+^{14}C_{14}(0.901)^{14}(1-0.901)^0\\\\=(14)(0.901)^{13}(0.099)+(1)(0.901)^{14}\ \ [\because\ ^nC_n=1\ \&\ ^nC_{n-1}=n ]\\\\\approx0.3574+0.2324=0.5898$

∴ The required probability = 0.5898

5 0

1 year ago

A jar contains 0.25 liter of apple juice and 0.30 liter of grape juice. Melissa poured 0.75 liter of pineapple juice into the ja

Advocard [28]

Part A: [P + (A + G) - M

Part B: [0.75 + (0.25 + 0.30) - 0.20] = 1.1

8 0

2 years ago

Nuri joins a game for a car. The rule is that Nuri pick one key from box either A, B, or C. A box has two keys but only one can

jek_recluse [69]

Answer:

0.278

Step-by-step explanation:

Given that Nuri joins a game for a car. The rule is that Nuri pick one key from box either A, B, or C. A box has two keys but only one can be used. B box has three keys but only one can be used. C box has two keys but none of them can be used.

Each box is equally likely to be selected.

In other words

$P(A) = P(B) = P(C)=\frac{1}{3}$

If A is selected then probability of winning is using the correct key out of two keys i.e. 0.5

If B is selected then probability of winning is using the correct key out of three keys i.e. 0.333

If c is selected then probability of winning is using the correct key out of two keys i.e. 0.00

So the probability that Nuri can win the car

= $\frac{1}{3} *0.5+\frac{1}{3} *0.333+\frac{1}{3} *0\\= 0.278$

6 0

1 year ago

a graduated cylinder is filled to 41.5 ml with water and a piece of granite is placed in the cylinder displacing the level to 47

mixas84 [53]

Because you went from 41.5 to 47.6 when the granite was placed in the graduated cylinder, using displacement:

47.6 - 41.5 = 6.1

I don't know about the cubic centimeters part, I don't know for sure if mL and cubic cm are the same, sorry. :(

8 0

1 year ago