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3 years ago

13

Dan sold 150,000 candies at the fair. he sold 10 times more candy this year than last year. how many candies did Dan sell last y

ear?

2 answers:

AlexFokin [52]3 years ago

7 0

Dan sold 15,000 candies last year at the fair

Rainbow [258]3 years ago

6 0

What is 150thousand times ten?

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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.

<em>Let </em> $\bar X$ <em> = sample average salary</em>

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)

<em>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.</em>

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

4 0

2 years ago

Find the absolute value of 2 and 2 3rd and negative 9 over 4

gayaneshka [121]

we are given

absolute value of 2 and 2 3rd and negative 9 over 4

Firstly , we need write it in terms of expression

2 and 2 3rd is

$2+\frac{2}{3}$

so, we get as

$|2+\frac{2}{3} -\frac{9}{4} |$

now, we can simplify it

Firstly , we will find common denominator

$|\frac{2*12}{1*12} +\frac{2*4}{3*4} -\frac{9*3}{4*3} |$

$|\frac{24}{12} +\frac{8}{12} -\frac{27}{12} |$

we can see that

all denominators are same

so, we can combine numerators

$=|\frac{24+8-27}{12} |$

$=|\frac{5}{12} |$

$=\frac{5}{12}$ ..............Answer

6 0

3 years ago

Is 3/20 larger then 2

aleksley [76]

To find out is 3/20 is larger than 2 we will convert 3/20 to a decimal. Lets do it:-

3/20
3 ÷ 20 = 0.15

Is 0.15 larger than 2?

No, 0.15 is not larger than 2.

So, 3/20 isn't larger than 2.

Hope I helped ya!! xD

8 0

2 years ago

Read 2 more answers

Can someone help 30 points!

irina1246 [14]

We cannot solve for a numerical value of p, but rather set the equation equal to p.

w = 4/(p+2)

(Multiply both sides by the denominator)

w(p+2) = 4

(divide both sides by w)

p+2 = 4/w

(Subtract 2 from both sides) (also 2/1)

p = (4-2)/w

p = 2/w

3 0

2 years ago

19.214÷13 <br> pls someone help me

slava [35]

Answer:

1.478

Step-by-step explanation:

5 0

2 years ago