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3 years ago

8

Suppose that a test is standardized (normally distributed) with a mean of 75 and a standard deviation of 11. what is the probabi

lity that an individual will have a test score greater than 80

1 answer:

uranmaximum [27]3 years ago

5 0

P(x > 80) ≈ 32.47%

A suitable calculator can help you figure this.

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Score: 4 of 8 pts

pochemuha

Answer:

10.5 ft high

5.3 ft horizontally

Step-by-step explanation:

The equation can be written in vertex form to answer these questions.

f(x) = -0.2(x² -10.5x) +5

f(x) = -0.2(x² -10.5x +5.25²) +5 +0.2(5.25²)

f(x) = -0.2(x -5.25)² +10.5125

The vertex of the travel path is (5.25, 10.5125).

The maximum height is 10.5 feet, which occurs 5.3 feet (horizontally) from the point of release.

7 0

3 years ago

11. A physical therapist is using a wide elastic band to help a

Vinvika [58]

Answer:

(C) 5 decimeters

Step-by-step explanation:

A therapist is using an elastic band to strengthen a patient's knee after surgery.

The patient's knee can tolerate a maximum of 20 kg i.e. 20000 gm weight of force.

If the elastic band provides 40 gm of weight force for every millimeter of its extension.

then this can be extended up to $\frac{20000}{40} =500$ millimeters, which is equal to 5 decimeters.

Therefore, option c is correct. (Answer)

7 0

3 years ago

Make a stem-and-leaf plot for the following data. 59, 38, 33, 26, 44, 35, 32, 47, 45, 24, 27, 46, 34, 30, 36

aliina [53]

Stem : leaf :
2 4,6,7
3 0,2,3,4,5,6,8
4 4,5,6,7
5 9

7 0

3 years ago

Read 2 more answers

HELP PLZ; 20 POINTS

Minchanka [31]

(a) 1A 2A 3A 1B 2B 3B (6 possible outcomes

(b) Possible outcomes for A or 1 are 1A 2A 3A and 1B ( 4 outcomes) therefore:
Probabilty = 4/6 = 2/3

6 0

3 years ago

Integrate the following problem:

vazorg [7]

Answer:

$\displaystyle \frac{2 \cdot sin2x-cos2x}{5e^x} + C$

Step-by-step explanation:

The integration by parts formula is: $\displaystyle \int udv = uv - \int vdu$

Let's find u, du, dv, and v for $\displaystyle \int e^-^x \cdot cos2x \ dx$ .

$u=e^-^x$
$du=-e^-^x dx$

$dv=cos2x \ dx$
$v= \frac{sin2x}{2}$

Plug these values into the IBP formula:

$\displaystyle \int e^-^x \cdot cos2x \ dx = e^-^x \cdot \frac{sin2x}{2} - \int \frac{sin2x}{2} \cdot -e^-^x dx$
$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \int \frac{sin2x}{2} \cdot -e^-^x dx$

Now let's evaluate the integral $\displaystyle \int \frac{sin2x}{2} \cdot -e^-^x dx$ .

Let's find u, du, dv, and v for this integral:

$u=-e^-^x$
$du=e^-^x dx$
$dv=\frac{sin2x}{2} dx$
$v=\frac{-cos2x}{4}$

Plug these values into the IBP formula:

$\displaystyle \int -e^-^x \cdot \frac{sin2x}{x}dx = -e^-^x \cdot \frac{-cos2x}{4} - \int \frac{-cos2x}{4}\cdot e^-^x dx$

Factor 1/4 out of the integral and we are left with the exact same integral from the question.

$\displaystyle \int -e^-^x \cdot \frac{sin2x}{x}dx = -e^-^x \cdot \frac{-cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx$

Let's substitute this back into the first IBP equation.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \Big [ -e^-^x \cdot \frac{-cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx \Big ]$

Simplify inside the brackets.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \Big [ \frac{e^-^x \cdot cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx \Big ]$

Distribute the negative sign into the parentheses.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4} - \frac{1}{4} \int cos2x \cdot e^-^x dx$

Add the like term to the left side.

$\displaystyle \int e^-^x \cdot cos2x \ dx + \frac{1}{4} \int cos2x \cdot e^-^x dx= \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4}$
$\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4}$

Make the fractions have common denominators.

$\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x}{4} - \frac{e^-^x \cdot cos2x}{4}$

Simplify this equation.

$\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{4}$

Multiply the right side by the reciprocal of 5/4.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{4} \cdot \frac{4}{5}$

The 4's cancel out and we are left with:

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{5}$

Factor $e^-^x$ out of the numerator.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x(2 \cdot sin2x-cos2x)}{5}$

Simplify this by using exponential properties.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2 \cdot sin2x-cos2x}{5e^x}$

The final answer is $\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2 \cdot sin2x-cos2x}{5e^x} + C$ .

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3 years ago

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