What is the y- intercept of the line whose equation is y= 7x+5

Suppose that a basketball player different from the

Volgvan

Answer:

a. Mean = 6

Variance = 2.4

Standard Deviation = 1.55

b. P(X=16) = 0.124

Step-by-step explanation:

Given

n = Total shots = 10

p = Probability of success = 60%

p = 60/100

p= 0.6

q = Probability of failure

q = 1-p

q = 1 - 0.6

q = 0.4

a.

Mean = np

Mean = 10 * 0.6

Mean = 6

Variance = npq

Variance = 10 * 0.6 * 0.4

Variance = 2.4

Standard Deviation = √Variance

Standard Deviation = √2.4

Standard Deviation = 1.549193338482966

Standard Deviation = 1.55 --------- approximated

b.

We have X = 16

x = 10

Assume that the events "success" on the various throws are independent.

The 10th success came on the 16th attempt

So, the player had exactly 10 successes and 6 failures on 16th trial

So Probability = nCr 0.6^10 * 0.4^6

Where n = 15 and r = 9 (number of attempts and success before the 16th trial)

15C9 * 0.6^10 * 0.4^6

= 5005 * 0.0060466176 * 0.004096

= 0.123958563176448

= 0.124 ------ Approximated

6 0

3 years ago

PLEASE NO LINKS.

Kamila [148]

Answer:

Step-by-step explanation:

Sustituye la variable

y

con

|

x

−

3

|

+

|

x

+

2

|

−

|

x

−

5

|

en la expresión.

s

i

−

2

3 0

3 years ago

The following data represent the age (in weeks) at which 10 randomly selected babies crawled for the first time. 43, 44, 42, 31,

Studentka2010 [4]

Answer:

$42-1.833\frac{6.325}{\sqrt{10}}=38.334$

$42+1.833\frac{6.325}{\sqrt{10}}=45.666$

The 90% confidence interval is given by $38.334 \leq \mu \leq 45.666$

Step-by-step explanation:

Notation

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=42$

The sample deviation calculated $s=6.325$

The degrees of freedom are given by:

$df=n-1=10-1=9$

The Confidence level is 0.90 or 90%, the significance is $\alpha=0.1$ and $\alpha/2 =0.05$ , and the critical value would be $t_{\alpha/2}=1.833$

Replacing we got:

$42-1.833\frac{6.325}{\sqrt{10}}=38.334$

$42+1.833\frac{6.325}{\sqrt{10}}=45.666$

The 90% confidence interval is given by $38.334 \leq \mu \leq 45.666$

5 0

4 years ago

#2

mars1129 [50]

Answer:

a) 16.5 inches

b) 1.64 radian

c) 22.3feet

Step-by-step explanation:

$s = rθ \:$

a) r = 3in

θ = 5.5rad

s = 3 × 5.5

= 16.5in

b) s = 6.4in

r = 3.9in

6.4 = 3.9 × θ

θ = 1.64rad (3sf)

$arc= \frac{θ}{360} \times 2 \: \pi \: r$

c) θ = 126°

arc = 49ft

49 = (126÷360)×2×3.142×r

r = 49 ÷ 2.1994

= 22.3ft (3sf)

(Correct me if i am wrong)

8 0

4 years ago

Please help me out with this please

Natali [406]

Answer:

Step-by-step explanation:

We have 2 equations represented by the lines on the graph

5x + 4y = 20

2x - 6y = 12

To plot the first equation on the graph, we a assume different points

4y = 20 - 5x

y = (20-5x)/4

y = 5 - 5x/4

If x =0, y = 5

If x = 2, y = 2.5

If x = 4, y = 0,

These points corresponds to the first line that cuts the positive y axis.

The first line that cuts the positive y axis is represented by the equation,

5x + 4y = 20

Since the left region of the line representing equation is shaded, the unshaded side represents

5x + 4y lesser than 20

To plot the second equation on the graph, we a assume different points

-6y = 12-2x

y = (2x-12)/6 = x/3 - 2

if x= 0, y = -2

If x = 3,y = 1

These points corresponds to the second line that cuts the negative y axis.

The second line that cuts the negative y axis is represented by the equation,

2x -6y = 12

Since the downward region of the line representing the equation is shaded, the unshaded side represents

2x -6y greater than 12

8 0

3 years ago