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3 years ago

Carl and Beneta are playing a game using the spinner below

Mathematics

1 answer:

Molodets [167]3 years ago

3 0

A) Carl's claim is incorrect since the probability of the spinner landing on a 6, 7, or 8 is 3/8, which is only 37.5% and cannot be described as likely.

b) For Benata there are many situations for which this could work, but one example for event X would be the spinner landing on a number less than or equal to 6.

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Karen Sullivan delivers newspapers for the Tribune dispatch. She receives 18.2 cents per paper, six days a week (for the daily p

mario62 [17]

Answer:

About $241.11

Step-by-step explanation:

So, Karen receives 18.2 cents per paper.

She delivers 124 paper per day.

In other words, on days other than Sunday, she will make a total of:

$18.2(124)=2256.8\text{ cents}$

On Sunday, each paper is sold for $0.70 or 70 cents. She also sells 151 Sunday papers. Thus, on a Sunday, she will make a total of:

$70(151)=10570\text{ cents}$

Therefore, in one week, she will do the first equation six times and the Sunday equation once. Thus, her total pay will be:

$6(2256.8)+10570=24110.8\text{ cents}\approx\$241.11$

4 0

2 years ago

Solve the equation<br><br> (-7) (-5)

Tanzania [10]

The answer is 35 i think but i am not sure

6 0

3 years ago

Write 61/4℅ as a decimal

marta [7]

61/4 % converted into a decimal is 15.25

3 0

3 years ago

he amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and s

sp2606 [1]

Complete question:

He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is

a) less than 8 minutes

b) between 8 and 9 minutes

c) less than 7.5 minutes

Answer:

a) 0.0708

b) 0.9291

c) 0.0000

Step-by-step explanation:

Given:

n = 47

u = 8.3 mins

s.d = 1.4 mins

a) Less than 8 minutes:

$P(X$

P(X' < 8) = P(Z< - 1.47)

Using the normal distribution table:

NORMSDIST(-1.47)

= 0.0708

b) between 8 and 9 minutes:

P(8< X' <9) = $[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]$

= P(-1.47 <Z< 6.366)

= P( Z< 6.366) - P(Z< -1.47)

Using normal distribution table,

$NORMSDIST(6.366)-NORMSDIST(-1.47)$

0.9999 - 0.0708

= 0.9291

c) Less than 7.5 minutes:

P(X'<7.5) = $P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]$

P(X' < 7.5) = P(Z< -3.92)

NORMSDIST (-3.92)

= 0.0000

3 0

2 years ago

Show all work please!!!!!!

sergij07 [2.7K]

The midpoint formula is:
(point1 + point2)/2

1) (-2 + 12)/2 = 5

2) (-1 +5)/2 = 2
    (0+2)/2 = 1
     (2,1)

3) Sqrt of (x2-x1)^2 + (y2 - y1)^2
     sqrt of (7 -1)^2 + (-7 - 1)^2
     sqrt of (6)^2 + (-8)^2
     sqrt of 36 + 64
     sqrt of 100
     10

Hope this helps!

7 0

3 years ago