Eleven is 20% of a0.

What is an example of a square root function? In (,) form

RUDIKE [14]

Step-by-step explanation:

The parent function of the functions of the form f(x)=√x−a+b is f(x)=√x . Note that the domain of f(x)=√x is x≥0 and the range is y≥0 . The graph of f(x)=√x−a+b can be obtained by translating the graph of f(x)=√x to a units to the right and then b units up.

I hope it's helpful!

5 0

3 years ago

The perimeter of Suzanne's rectangular garden is 40 yards. The length of the garden is 12 yards. What is the area of Suzanne's g

asambeis [7]

Answer:

96 sq yards

Step-by-step explanation:

40-24 is 16, and 16/2 is 8. 8 is the second length, so 8 times 12 is 96 which is the area

5 0

3 years ago

A study was conducted on 64 female college athletes. The researcher collected data on a number of variables including percent bo

Leni [432]

Answer:

Hello please your question is in-complete attached is the complete question

degree of freedom = -62.90 ( e )

Step-by-step explanation:

The formula for calculating the F-statistic/test statistic is

test - statistic = Coef ( LBW) / SE Coef ( LBW )

= -0.72399 / 0.01151

= - 62.90

the degree of freedom the F-statistic has = -62.90

F-statistic test is any statistical test in which the test statistic has an F-distribution under the null hypothesis. the value of the test can be gotten from running an ANOVA test or regression analysis on the statistical models

4 0

3 years ago

the length of a tangent from a point a at distance 5 cm from the centre of the circle is 4 cm find the radius of the circle

Semenov [28]

The radius of the circle is 3 cm.

<u>Step-by-step explanation:</u>

Refer the attached diagram, the circle with centre O. In that given, AB is tangent given as 4 cm and distance of point from the circle, OA = 5 cm

As AB is tangent, OB (radius of circle) is perpendicular to AB (tangent at any point of circle). Therefore the angle of OBA is 90 degree.

Also, triangle OAB is a right angled triangle (refer attached diagram). By using Pythagoras theorem in right angled triangle,

$(\text {Hypotenuse})^{2}=(\text {Height})^{2}+(\text {Base})^{2}$

$(O A)^{2}=(O B)^{2}+(A B)^{2}$

Substitute the given values in the above expression, we get

$(5)^{2}=(O B)^{2}+(4)^{2}$

$(O B)^{2}=25-16=9$

Taking square root on both side, we get

Radius of the circle, OB = 3 cm

5 0

3 years ago

Consider a Poisson distribution with μ = 6.

bearhunter [10]

Answer:

a) $P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

b) f(2) = 0.04462

c) f(1) = 0.01487

d) $P(X \geq 2) = 0.93803$

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of successes

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

In this question:

$\mu = 6$

a. Write the appropriate Poisson probability function.

Considering $\mu = 6$

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

b. Compute f (2).

This is P(X = 2). So

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

So f(2) = 0.04462

c. Compute f (1).

This is P(X = 1). So

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

So f(1) = 0.01487.

d. Compute P(x≥2)

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-6}*6^{0}}{(0)!} = 0.00248$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

Then

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00248 + 0.01487 + 0.04462 = 0.06197$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.06197 = 0.93803$

So

$P(X \geq 2) = 0.93803$

5 0

2 years ago