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1 year ago

A diagram of a lot of land the scale on the diagram is 1 cm represents 2.5 yards what is the area of the actual lot of land

Mathematics

1 answer:

netineya [11]1 year ago

5 0

In order to determine the area of the lot of land in squared yards, proceed as follow:

Convert the dimensions of the rectangle to yards, by using the given conversion factor:

$\begin{gathered} 7\operatorname{cm}\cdot\frac{2.5yd}{1\operatorname{cm}}=17.5yd \\ 15\operatorname{cm}\cdot\frac{2.5yd}{1\operatorname{cm}}=37.5yd \end{gathered}$

Next, consider that the area of the lot is the area of a rectangle. Replace the previous values of the length and the width of the rectangle:

$A=(17.5yd)(37.5yd)=656.25yd^2$

Hence, the area of the lot of land is 656.25 sq yd

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) One year, professional sports players salaries averaged $1.5 million with a standard deviation of $0.9 million. Suppose a samp

Nutka1998 [239]

Answer:

Probability that the average salary of the 400 players exceeded $1.1 million is 0.99999.

Step-by-step explanation:

We are given that one year, professional sports players salaries averaged $1.5 million with a standard deviation of $0.9 million.

Suppose a sample of 400 major league players was taken.

Let $\bar X$ = sample average salary

The z-score probability distribution for sample mean is given by;

Z = $\frac{ \bar X -\mu}{{\frac{\sigma}{\sqrt{n} } }} }$ ~ N(0,1)

where, $\mu$ = mean salary = $1.5 million

$\sigma$ = standard deviation = $0.9 million

n = sample of players = 400

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, probability that the average salary of the 400 players exceeded $1.1 million is given by = P( $\bar X$ > $1.1 million)

P( $\bar X$ > $1.1 million) = P( $\frac{ \bar X -\mu}{{\frac{\sigma}{\sqrt{n} } }} }$ > $\frac{ 1.1-1.5}{{\frac{0.9}{\sqrt{400} } }} }$ ) = P(Z > -8.89) = P(Z < 8.89)

Now, in the z table the maximum value of which probability area is given is for critical value of x = 4.40 as 0.99999. So, we can assume that the above probability also has an area of 0.99999 or nearly close to 1.

Therefore, probability that the average salary of the 400 players exceeded $1.1 million is 0.99999.

4 0

3 years ago

Anyone good with algebra 2??

sweet [91]

A discontinuity is a point that cannot exist because the x-coordinate would cause a problem in the equation. If you have a polynomial in the denominator, you must find which values of x would cause the polynomial in the denominator to evaluate to zero. Since division by zero is undefined, that would cause a discontinuity.

Let's look at your function.

$f(x) = \dfrac{x^2 - 16}{6x - 24}$

$x^2 - 16$ is in the numerator. It is defined for every value of x. There is no problem there.

$6x - 24$ is in the denominator. This is a function defined for every value of x, but since it is in the denominator, we must exclude the x-value that would cause this polynomial to evaluate to zero.

We set it equal to zero and solve the equation for x.

$6x - 24 = 0$

$6x = 24$

$x = 4$

For x = 0, the denominator has a value of zero, so at this point there is a discontinuity in function f(x).

The answer is:

$x \ne 4$

6 0

3 years ago

-4|x-11|=-16 solve for x in the equation

BigorU [14]

Answer:

x = 7 and 15

Step-by-step explanation:

Divide both sides by -4...

| x - 11 | = 4

because

| -4 | = 4 and

| 4 | = 4

x - 11 = -4 and x - 11 = 4

solve for x in both equations.

x = 7 and x = 15

5 0

3 years ago

Linear or nonlinear ?

timurjin [86]

Nonlinear is correct the points are scattered across the graph :)

5 0

2 years ago

In a study, researchers wanted to measure the effect of alcohol in the development of the hippocampal region in adolescents. The

lbvjy [14]

Answer:

Kindly check explanation

Step-by-step explanation:

Given :

Sample size, n = 30

Tcritical value = 2.045

Null hypothesis :

H0: μ = 9.08

Alternative hypothesis :

H1: μ≠ 9.08

Sample mean, m = 8.25

Samole standard deviation, s = 1.67

Test statistic : (m - μ) ÷ s/sqrt(n)

Test statistic : (8.25 - 9.08) ÷ 1.67/sqrt(30)

Test statistic : - 0.83 ÷ 0.3048988

Test statistic : - 2.722

Tstatistic = - 2.722

Decision region :

Reject Null ; if

Tstatistic < Tcritical

Tcritical : - 2.045

-2.722 < - 2.045 ; We reject the Null

Using the α - level (confidence interval) 0.05

The Pvalue for the data from Tstatistic calculator:

df = n - 1 =. 30 - 1 = 29

Pvalue = 0.0108

Reject H0 if :

Pvalue < α

0.0108 < 0.05 ; Hence, we reject the Null

4 0

2 years ago