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BlackZzzverrR [31]

2 years ago

5

On a game show, contestants shoot a foam ball toward a target. The table includes points along OnePath the ball can take to hit

the target where X is the time that has passed since the ball was launch and why is the height at this time.

1 answer:

igor_vitrenko [27]2 years ago

7 0

Answer:

42

Step-by-step explanation:

You might be interested in

What is the exact value of the cos F?

Galina-37 [17]

Answer:

cos(F) = 9/41

Step-by-step explanation:

The triangles are similar, so you know that ...

... cos(D) = cos(A) = 40/41.

From trig relations, you know ...

... cos(F) = sin(D)

and

... sin(D)² +cos(D)² = 1

So ...

... cos(F) = sin(D) = √(1 -cos(D)²) = √(1 -(40/41)²) = √(81/1681)

... cos(F) = 9/41

_____

The ratio for cos(A) tells you that you can consider AB=40, AC=41. Then, using the Pythagorean theorem, you can find BC = √(41² -40²) = √81 = 9.

From the definition of the cosine, you know cos(C) = BC/AC = 9/41. Because the triangles are similar, you know

... cos(F) = cos(C) = 9/41

7 0

2 years ago

Production of passenger cars in Japan increased from 3.94 million in 1999 to 6.74 million in 2009. What is the geometric mean an

vovangra [49]

Let's assume initial population is in 1999

so, production of passenger cars in Japan is 3.94 million in 1999

so, P=3.94 million

and the production of passenger cars in Japan is 6.74 million in 2009

so, A=6.74 million in t=2009-1999=10 years

now, we can use formula

$A=P(1+r)^t$

here , r is interest rate

so, we can plug values

$6.74=3.94(1+r)^{10}$

now, we can solve for

$\left(1+r\right)^{10}=\frac{337}{197}$

$r=0.05516$

so,

the geometric mean annual percent increase is 5.516%............Answer

7 0

2 years ago

The extract of a plant native to Taiwan has been tested as a possible treatment for Leukemia. One of the chemical compounds prod

True [87]

Answer:

a) 57.35%

b) 99.99%

c) 68.27%

Step-by-step explanation:

When we have a random variable X that is normally distributed with mean $\large\bf \mu$ and standard deviation $\large\bf \sigma$ , then

The probability that the random variable has a value less than a, P(X < a) = P(X ≤ a) is the area under the normal curve with mean $\large\bf \mu$ and standard deviation $\large\bf \sigma$ to the left of a.

The probability that the random variable has a value greater than b, P(X > b) = P(X ≥ b) is the area under the normal curve with mean $\large\bf \mu$ and standard deviation $\large\bf \sigma$ to the right of b.

The probability that the random variable has a value between a and b, P(a < X < b) = P(a ≤ X ≤ b) = P(a < X ≤ b)= P(a ≤ X < b) is the area under the normal curve with mean $\large\bf \mu$ and standard deviation $\large\bf \sigma$ between a and b.

In this case, the random variable is the collagen amount found in the extract of the plant. The mean is 63 g/ml and the standard deviation is 5.4 g/ml

(a) What is the probability that the amount of collagen is greater than 62 grams per mililiter?

As we have seen, we need to find the area under the normal curve with mean 63 and standard deviation 5.4 to the right of 62 (see picture).

You can find this value easily with a calculator or a spreadsheet. If you prefer the old-style, then you have to standardize the values and look up in a table.

<em>If you have access to Excel or OpenOffice Calc, you can find this value by introducing the formula: </em>

<em>1- NORMDIST(62,63,5.4,1) in Excel </em>

<em>1 - NORMDIST(62;63;5.4;1) in OpenOffice Calc </em>

<em>and we will get a value of 0.5735 or 57.35% </em>

(b) What is the probability that the amount of collagen is less than 90 grams per mililiter?

Now we want the area to the left of 90

<em>NORMDIST(90,63,5.4,1) in Excel </em>

<em>NORMDIST(90;63;5.4;1) in OpenOffice Calc </em>

You will get a value of 0.9999 or 99.99%

(c) What percentage of compounds formed from the extract of this plant fall within 1 standard deviations of the mean?

You can use either the rule that 68.27% of the data falls between $\large\bf \mu -\sigma$ and $\large\bf \mu +\sigma$ or compute area between 63 - 5.4 and 63 + 5.4, that is to say, the area between 57.6 and 68.4

<em>In Excel </em>

<em>NORMDIST(68.4,63,5.4,1) - NORMDIST(57.6,63,5.4,1) </em>

<em>In OpenOffice Calc </em>

<em>NORMDIST(68.4;63;5.4;1) - NORMDIST(57.6;63;5.4;1) </em>

In any case we get a value of 0.6827 or 68.27%

3 0

2 years ago

Which relationship has a zero slope?

Ivenika [448]

9514 1404 393

Answer:

A. A two column table with five rows. The first column, x, has the entries, negative 3, negative 1, 1, 3. The second column, y, has the entries, 2, 2, 2, 2

Step-by-step explanation:

The relationship that has constant y-values is the one with zero slope. That would be the first table.

6 0

2 years ago

The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100,

Gre4nikov [31]

Answer:

A.the type 1 error probability is $\mathbf{\alpha = 0.0244 }$

B. β = 0.0122

C. β = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

$\mathtt{H_o: \mu = 100}$

$\mathtt{H_1: \mu \neq 100}$

A. If the acceptance region is defined as $98.5 < \overline x > 101.5$ , find the type I error probability $\alpha$ .

Assuming the critical region lies within $\overline x < 98.5$ or $\overline x > 101.5$ , for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is $\mu = 100$

∴

$\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}$

$\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}$

when $\mu = 100$

$\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }$

$\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }$

$\mathtt{\alpha = P ( Z 2.25) }$

$\mathtt{\alpha = P ( Z$

From the standard normal distribution tables

$\mathtt{\alpha = 0.0122+( 1- 0.9878) })$

$\mathtt{\alpha = 0.0122+( 0.0122) })$

$\mathbf{\alpha = 0.0244 }$

Thus, the type 1 error probability is $\mathbf{\alpha = 0.0244 }$

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis $\mathtt{H_o}$

Thus;

β = P( type II error) - P( fail to reject $\mathtt{H_o}$ )

∴

$\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }$

Given that $\mu = 103$

$\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }$

$\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }$

$\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }$

$\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}$

From standard normal distribution table

β = 0.0122 - 0.0000

β = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

$\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }$

Given that $\mu = 105$

$\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }$

$\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }$

$\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }$

$\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}$

From standard normal distribution table

β = 0.0000 - 0.0000

β = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

8 0

2 years ago