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2 years ago

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Which of the following probabilities is the greatest for a standard normal distribution? P (negative 1.5 less-than-or-equal-to z

less-than-or-equal-to negative 0.5) P (negative 0.5 less-than-or-equal-to z less-than-or-equal-to 0.5) P (0.5 less-than-or-equal-to z less-than-or-equal-to 1.5) P (1.5 less-than-or-equal-to z less-than-or-equal-to 2.5)

1 answer:

hichkok12 [17]2 years ago

7 0

Answer:

15

Step-by-step explanation:

I hope that answered your question

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Geraldo wants to start a group bike ride on Sunday mornings for students at his school. He asked 30 students in the lunchroom ho

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4 is the answer because there are only 4 numbers greater than 10.Those are the people who ride bikes ore than 10 times.

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2 years ago

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The functions $f$ and $g$ are defined as follows: \[f(x) = \sqrt{\dfrac{x+1}{x-1}}\quad\text{and}\quad g(x) = \dfrac{\sqrt{x+1}}

CaHeK987 [17]

Answer:

The answer is "domain and range are different".

Step-by-step explanation:

Given:

$f(x) = \sqrt{\frac{x+1}{x-1}}\\\\g(x) =\frac{\sqrt{x+1}}{\sqrt{x-1}}\\$

Solve f(x) to find domain and range:

for element x: R:

⇒ $x \leq -1 \ \ \ and \ \ \ x > 1$ range:

$for \ \ f(x) \times element : 0 \leq f(x)< 1 \ \ \ or \ \ f(x)>1$

$g(x) =\frac{\sqrt{x+1}}{\sqrt{x-1}}\\$

Solve for domain:

⇒ $\ when \ x \ element \ R: \ \ \ x>1$

Solve for range:

⇒ $g(x) \times element$ R : $g(x)>1$

So, the value of the method f(x) and g(x) (range and domain) were different.

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2 years ago

An online furniture store sells chairs for $200 each and tables for $600 each. Every day, the store can ship a maximum of 32 pie

wolverine [178]

Answer:

If the minimum 13 chair were sold , then the minimum number of Table sold is 14

Step-by-step explanation:

Given as :

The cost of each chair = $ 200

The cost of each table = $ 600

The total number of furniture sold per day = 32

The minimum amount of selling per day = $ 12000

Let the total number of chair = C

The total number of table = T

So , according to question

C + T = 32 .......1

200 C + 600 T = 12000 .......2

Solving eq 1 and 2

200 C + 600 T = 12000

200 × ( C + T ) = 32 × 200

I.e 200 C + 200 T = 6400

or, ( 200 C + 600 T ) - ( 200 C + 200 T ) = 12,000 - 6400

Or, 400 T = 5600

∴ T = $\frac{5600}{400}$

I.e T = 14

Put The value of T in eq 2

So, 200 C + 600 × 14 = 12000

or , 200 C + 8400 = 12,000

Or, 200 C = 12000 - 8400

Or, 200 C = 3600

∴ C = $\frac{3600}{200}$

I.e C = 18

The number of chair sold is 18

If the number of chair sold is 13 ,

Then the min number of table sold = 200 × 18 + 600 T = 12000

i.e 600 T = 12000 - 3600

or, 600 T = 8400

∴ T = $\frac{8400}{600}$

I.e T = 14

Hence if the minimum 13 chair were sold , then the minimum number of Table sold is 14 . Answer

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2 years ago

A helicopter flies 8 km due north from A to B. It then flies 5 km from B to C and returns to A as shown in the figure. Angle ABC

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Please consider the attached graph.

We have been given that a helicopter flies 8 km due north from A to B. It then flies 5 km from B to C and returns to A as shown in the figure. The measure of angle ABC is 150°. We are asked to find the area of triangle ABC.

We will use trigonometric area formula to solve our given problem.

$A=\frac{1}{2}a\cdot c\cdot\text{sin}(b)$ , where angle b is angle between sides a and c.

For our given triangle $a=5,c=8$ and measure of angle b is 150 degrees.

$A=\frac{1}{2}(5)\cdot (8)\cdot\text{sin}(150^{\circ})$

$A=5\cdot 4\cdot(0.5)$

$A=10$

Therefore, the area of the triangle ABC is 10 square kilo-meters and option 'c' is the correct choice.

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2 years ago

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

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

<em />

<u>So, 90% confidence interval for the population proportion, p is ;</u>

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

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

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

= [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%.

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2 years ago