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fenix001 [56]
3 years ago
7

Holly bought 23 Tickets to the baseball game. She got a group rate that gave her $4 off the regular ticket price for each ticket

. The total cost was $184
What was the regular price for each ticket? PLEASE HELP
Mathematics
1 answer:
Ad libitum [116K]3 years ago
6 0
The answer is $12 per ticket.
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In a sample of 57 temperature readings taken from the freezer of a restaurant, the mean is 29.6 degrees and the population stand
Natalka [10]

Answer:

(29.14 ; 30.06)

Step-by-step explanation:

Given that'

Sample size (n) = 57

Mean (m) = 29.6

Population standard deviation (σ) = 2.7

Confidence interval = 80%

= (1 - 0.8) / 2 = 0.1

Mean ± z * σ/√n

Using the Z probability calculator : Z0. 1 = 1.28

Hence,

29.6 ± 1.28 * (2.7 / √57)

29.6 - (1.28 * 0.3576237) ; 29.6 + (1.28 * 0.3576237)

29.142241664 ; 30.057758336

(29.14 ; 30.06)

8 0
3 years ago
Please help! <br><br> ~I WILL MARK BRAINLIEST!~
likoan [24]
I’m pretty sure it is B
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6 0
2 years ago
Read 2 more answers
Lisa is mailing a letter to a friend. Postage costs $0.32 for the first ounce and $0.23 for the additional ounce. How heavy can
Advocard [28]

Answer: 8.3 ounces

Step-by-step explanation:

simple, there's a flat fee for 1 ounce, then 23 cents for each ounce following.

so $2 - $0.32 = $1.68

1.67 / 0.23 = 7.3 ounces

7.3 ounces + 1 ounces (from the flat fee we subtracted)

which equals 8.3

7 0
2 years ago
Samples of 20 parts from a metal punching process are selected every hour. Typically, 1% of the parts require rework. Let X deno
Roman55 [17]

Answer:

a) P(X>np+3\sqrt{np(1-p)}=0.017

b) P(x>1)=0.190

c) P(Y>1)=0.651

Step-by-step explanation:

This a binomial experiment where success is denoted by parts that need rework.

X ∼ B(n, p); n = 20; p = 0.01

The expected value of X is: E(X) = np =20×0.01= 0.2

The variance is: Var(X) = np(1 − p) = 0.2 × 0.99 = 0.198,

The standard deviation SD(X)= \sqrt{0.198} ≈ 0.445

a) P(X>np+3\sqrt{np(1-p)}=P(X>0.2+3×0.445)=P(X>1.535)=P(X≥2)

Probability function is given by:

\frac{n!}{x!(n-x)!} *p^x*(1-p)^{(n-x)}

P(X≥2)=1-P(X<2)=1-P(X=1)-P(X=0)= 1 - \frac{20!}{1!(20-1)!} *(0.01)^{1}*(1-0.01)^{(20-1)}-\frac{20!}{0!(20-0)!} *(0.01)^{0}*(1-0.01)^{(20-0)}

P(X≥2)=1-0.165-0.818=0.017

b) p=0.04

P(x>1)=P(x≥2)= 1 - P(x=1) - P(x=0)= 1 - \frac{20!}{0!(20-1)!} *(0.04)^{1}*(1-0.04)^{(20-1)} - \frac{20!}{0!(20-0)!} *(0.04)^{0}*(1-0.04)^{(20-0)}

P(x>1)= 1 - 0.368 - 0.442=0.190

c) In this case we consider p=0.19 (Probability that X exceeds 1)

In this experiment Y is the number of hours and n= 5 hours.

Then, we check the probability in each hour:

P(Y>1)=1- P(Y=0)

P(Y=0)=\frac{5!}{0!(5-0)!} *(0.19)^{0}*(1-0.19)^{(5-0)}=0.349

P(Y>1)=1-0.349=0.651

3 0
3 years ago
Which angle is supplementary to angle # 3?
stealth61 [152]

Check the picture below.

4 0
2 years ago
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