Can someone help me I'm doing my packet and the quest is hard

An auction company sold 15 handheld computers last month. the final prices (in dollars) at which the items sold are as follows.

matrenka [14]

Hi,

a. construct a stem and leaf plot

Ans

20: 5

21: 3 7

22: 9 9 9 9

23: 1 6

24: 5 7 7 8 8

25: 2

b. summarize what this plot tells you about the data

Ans

This plot tells that most numbers of computers were sold at prices in the range of $220-229 and $240-249. For other prices, sales were lower.

Hoped I Helped

5 0

4 years ago

Read 2 more answers

PLS SOLVE THIS PAGE ASAP

d1i1m1o1n [39]

Y=mx+b
m is the slope rise/run and b is the y intercept plot the y intercept and from that point use the slope to find the next points. does that help any

3 0

3 years ago

The phrase “y less than 7” translates to an equation.<br><br> True or False?

defon

Answer:

False, it translates to an inequality.

8 0

4 years ago

In a binomial distribution, n = 8 and π=0.36. Find the probabilities of the following events. (Round your answers to 4 decimal p

skelet666 [1.2K]

Answer:

$\mathbf{P(X=5) =0.0888}$

P(x ≤ 5 ) = 0.9707

P ( x ≥ 6) = 0.0293

Step-by-step explanation:

The probability of a binomial mass distribution can be expressed with the formula:

$\mathtt{P(X=x) =(^{n}_{x} ) \ \pi^x \ (1-\pi)^{n-x}}$

$\mathtt{P(X=x) =(\dfrac{n!}{x!(n-x)!} ) \ \pi^x \ (1-\pi)^{n-x}}$

where;

n = 8 and π = 0.36

For x = 5

The probability $\mathtt{P(X=5) =(\dfrac{8!}{5!(8-5)!} ) \ 0.36^5 \ (1-0.36)^{8-5}}$

$\mathtt{P(X=5) =(\dfrac{8!}{5!(3)!} ) \ 0.36^5 \ (0.64)^{3}}$

$\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 \times 5!}{5!(3)!} ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 }{3 \times 2 \times 1} ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =({8 \times 7 } ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =0.0887645}$

$\mathbf{P(X=5) =0.0888}$ to 4 decimal places

b. x ≤ 5

The probability of P ( x ≤ 5) $\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})$

${P(x \leq 5) = ( \dfrac{8!}{0!(8!)} \times (0.36)^0 \times (1-0.36)^8 \ ) + \dfrac{8!}{1!(7!)} \times (0.36)^1 \times (1-0.36)^7 \ +$ $\dfrac{8!}{2!(6!)} \times (0.36)^2 \times (1-0.36)^6 \ + \dfrac{8!}{3!(5!)} \times (0.36)^3 \times (1-0.36)^5 + \dfrac{8!}{4!(4!)} \times (0.36)^4 \times (1-0.36)^4 \ + \dfrac{8!}{5!(3!)} \times (0.36)^5 \times (1-0.36)^3 \ )$