Write the following using index notation:

A newly hired basketball coach promised a high-paced attack that will put more points on the board than the team’s previously te

puteri [66]

Answer:

a. z = 2.00

Step-by-step explanation:

Hello!

The study variable is "Points per game of a high school team"

The hypothesis is that the average score per game is greater than before, so the parameter to test is the population mean (μ)

The hypothesis is:

H₀: μ ≤ 99

H₁: μ > 99

α: 0.01

There is no information about the variable distribution, I'll apply the Central Limit Theorem and approximate the sample mean (X[bar]) to normal since whether you use a Z or t-test, you need your variable to be at least approximately normal. Considering the sample size (n=36) I'd rather use a Z-test than a t-test.

The statistic value under the null hypothesis is:

Z= X[bar] - μ = 101 - 99 = 2

σ/√n 6/√36

I don't have σ, but since this is an approximation I can use the value of S instead.

I hope it helps!

7 0

3 years ago

-2x + 10 +4x=70 (what is X?)

Dovator [93]

Answer:

x = 30

Step-by-step explanation:

-2x + 10 +4x=70

-10 -10

-2x + 4x = 60

2x = 60

/2 /2

x = 30

8 0

3 years ago

Read 2 more answers

If <img src="https://tex.z-dn.net/?f=x%3D2z" id="TexFormula1" title="x=2z" alt="x=2z" align="absmiddle" class="latex-formula">,

Readme [11.4K]

Answer:

x = 1

z = 1/2

Step-by-step explanation:

z = 3/y

z = 3/6

z = 1/2

x = 2z

x = 2(1/2)

x = 1

8 0

2 years ago

Read 2 more answers

Angie read 22 pages during a 30-minute study hall. At this rate, how many pages would she read in 45 minutes

frosja888 [35]

45 minutes is equal to 1.5 times 30:

1.5*30=45

so she will also read 22*1.5 pages, which is 22+11=33

so she will read 33 pages

4 0

3 years ago

One of the earliest applications of the Poisson distribution was in analyzing incoming calls to a telephone switchboard. Analyst

grandymaker [24]

Answer:

(a) P (X = 0) = 0.0498.

(b) P (X > 5) = 0.084.

(c) P (X = 3) = 0.09.

(d) P (X ≤ 1) = 0.5578

Step-by-step explanation:

Let X = number of telephone calls.

The average number of calls per minute is, λ = 3.0.

The random variable X follows a Poisson distribution with parameter λ = 3.0.

The probability mass function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...$

(a)

Compute the probability of X = 0 as follows:

$P(X=0)=\frac{e^{-3}3^{0}}{0!}=\frac{0.0498\times1}{1}=0.0498$

Thus, the probability that there will be no calls during a one-minute interval is 0.0498.

(b)

If the operator is unable to handle the calls in any given minute, then this implies that the operator receives more than 5 calls in a minute.

Compute the probability of X > 5 as follows:

P (X > 5) = 1 - P (X ≤ 5)

$=1-\sum\limits^{5}_{x=0} { \frac{e^{-3}3^{x}}{x!}} \,\\=1-(0.0498+0.1494+0.2240+0.2240+0.1680+0.1008)\\=1-0.9160\\=0.084$

Thus, the probability that the operator will be unable to handle the calls in any one-minute period is 0.084.

(c)

The average number of calls in two minutes is, 2 × 3 = 6.

Compute the value of X = 3 as follows:

$P(X=3)=\frac{e^{-6}6^{3}}{3!}=\frac{0.0025\times216}{6}=0.09$

Thus, the probability that exactly three calls will arrive in a two-minute interval is 0.09.

(d)

The average number of calls in 30 seconds is, 3 ÷ 2 = 1.5.

Compute the probability of X ≤ 1 as follows:

P (X ≤ 1 ) = P (X = 0) + P (X = 1)

$=\frac{e^{-1.5}1.5^{0}}{0!}+\frac{e^{-1.5}1.5^{1}}{1!}\\=0.2231+0.3347\\=0.5578$

Thus, the probability that one or fewer calls will arrive in a 30-second interval is 0.5578.

5 0

3 years ago