Can somebody help me with this. 35 points! :)))))

The average score of all golfers for a particular course has a mean of 61 and a standard deviation of 3.5 . Suppose 49 golfers p

Nadya [2.5K]

Answer:

The probability that the average score of the 49 golfers exceeded 62 is 0.3897

Step-by-step explanation:

The average score of all golfers for a particular course has a mean of 61 and a standard deviation of 3.5

$\mu = 61$

$\sigma = 3.5$

We are supposed to find he probability that the average score of the 49 golfers exceeded 62.

Formula : $Z=\frac{x-\mu}{\sigma}$

$Z=\frac{62-61}{3.5}$

$Z=0.285$

Refer the z table for p value

p value = 0.6103

P(x>62)=1-P(x<62)=1-0.6103=0.3897

Hence the probability that the average score of the 49 golfers exceeded 62 is 0.3897

7 0

3 years ago

The circumference of a circle is 15ft. What is the length of the radius? Use 3.14 for pi. Round your answer to the nearest tenth

dybincka [34]

The circumference of the circle is determined by using the equation, C = 2πr. From this, the value of r may be solved by rearranging the equation to, r = C/2π. Substituting the given values,
r = 15 ft / (2 x 3.14)
The value of r is 2.3885 ft. Thus, the answer to the question is 2.4 ft.

3 0

3 years ago

An advertisement for a popular weight-loss clinic suggests that participants in its new diet program lose, on average, more than

Sedbober [7]

Testing the hypothesis, it is found that:

a)

The null hypothesis is: $H_0: \mu \leq 10$

The alternative hypothesis is: $H_1: \mu > 10$

b)

The critical value is: $t_c = 1.74$

The decision rule is:

If t < 1.74, we <u>do not reject</u> the null hypothesis.
If t > 1.74, we <u>reject</u> the null hypothesis.

c)

Since t = 1.41 < 1.74, we <u>do not reject the null hypothesis</u>, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

Item a:

At the null hypothesis, it is tested if the mean loss is of <u>at most 10 pounds</u>, that is:

$H_0: \mu \leq 10$

At the alternative hypothesis, it is tested if the mean loss is of <u>more than 10 pounds</u>, that is:

$H_1: \mu > 10$

Item b:

We are having a right-tailed test, as we are testing if the mean is more than a value, with a <u>significance level of 0.05</u> and 18 - 1 = <u>17 df.</u>

Hence, using a calculator for the t-distribution, the critical value is: $t_c = 1.74$ .

Hence, the decision rule is:

If t < 1.74, we <u>do not reject</u> the null hypothesis.
If t > 1.74, we <u>reject</u> the null hypothesis.

Item c:

We have the <u>standard deviation for the sample</u>, hence the t-distribution is used. The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.
$\mu$ is the value tested at the null hypothesis.
s is the standard deviation of the sample.
n is the sample size.

For this problem, we have that:

$\overline{x} = 10.8, \mu = 10, s = 2.4, n = 18$

Thus, the value of the test statistic is:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{10.8 - 10}{\frac{2.4}{\sqrt{18}}}$

$t = 1.41$

Since t = 1.41 < 1.74, we <u>do not reject the null hypothesis</u>, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

A similar problem is given at brainly.com/question/25147864

3 0

2 years ago

Find the domain for the particular solution to the differential equation dy dx equals the quotient of 3 times y and x , with ini

Svetach [21]

For this case we have the following difference equation:
$dy / dx = 3xy
$
Applying separable variables we have:
$dy / y = 3xdx
$
Integrating both sides we have:
$\int\ ({1/y}) \, dy = \int\ {3x} \, dx$
$ln (y) = (3/2) x ^ 2 + C
$
applying exponential to both sides:
$exp (ln (y)) = exp ((3/2) x ^ 2 + C)

 y = C * exp ((3/2) x ^ 2)$
For y (1) = 1 we have:
$C = 1 / (exp ((3/2) * 1 ^ 2))

C = 0.2$
Thus, the particular solution is:
$y = 0.2 * exp ((3/2) x ^ 2)
$
Whose domain is all real.
Answer:
y = 0.2 * exp ((3/2) x ^ 2)
Domain: all real numbers

6 0

3 years ago

Geometry Pythagorean theorem

lara31 [8.8K]

Answer:

Below!

Step-by-step explanation:

Using Pythagoras theorem, I will solve all of the problems.

<u>Question 9:</u>

10² = 6² + x²
=> 100 = 36 + x²
=> 100 - 36 = x²
=> 64 = x²
=> x = 8

<u>Question 10:</u>

26² = 24² + x²
=> 676 = 576 + x²
=> 676 - 576 = x²
=> 100 = x²
=> x = 10

<u>Question 11:</u>

15² = 12² + x²
=> 225 = 144 + x²
=> 225 - 144 = x²
=> 81 = x²
=> x = 9

<u>Question 12:</u>

x² = 8² + 12²
=> x² = 64 + 144
=> x² = 208
=> x = √208
=> x = 14.2 (Rounded)

<u>Question 13:</u>

7² = 2² + x²
=> 49 = 4 + x²
=> 49 - 4 = x²
=> 45 = x²
=> x = √45
=> x = 6.7 (Rounded)

<u>Question 14</u>

First, let's solve for the variable x using Pythagoras theorem.

=> 5² = 3² + x²
=> 25 = 9 + x²
=> 16 = x²
=> x = 4 units

Now, let's solve for the variable y using Pythagoras theorem.

(3 + 6)² = 5² + y²
=> (9)² = 25 + y²
=> 81 = 25 + y²
=> y² = 56
=> y = √56
=> y = 7.5 (Rounded) units

Answers (Nearest tenth):

x = 4 units
y = 7.5 units

<u>Question 15:</u>

First, let's find the value of the variable y using Pythagoras theorem.

8² = 6² + y²
=> 64 = 36 + y²
=> 28 = y²
=> y = √28
=> y = 5.3 (Rounded) units

Now, let's find the value of the variable x using multiplication.

x = 2y
=> x = 2(5.3)
=> x = 10.6 units

Answer (Nearest tenth)

x = 10.6 units
y = 5.3 units

5 0

2 years ago