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ipn [44]
3 years ago
10

The Wolfpack has won 11 games and lost 9 games this season. What percent of its games did the Wolfpack win?

Mathematics
2 answers:
defon3 years ago
8 0
First, you need to find the total amount of games played by adding 11 and 9
11 + 9 = 20
Now you have to divide 11 by 20. If you wanted to find the percent of games lost, you would divide 9 by 20.
11 \div 20
answer: 55%
Pavel [41]3 years ago
5 0

Answer:

55%

Step-by-step explanation:

Given data

  • Games won: 11 games
  • Games lost: 9 games

The total number of games is the sum of the games won and the games lost.

Total games = 11 games + 9 games = 20 games

The percent of games that the Wolfpack won is:

(Games won/Total games) × 100% = (11 games/20 games) × 100% = 55%

The Wolfpack won 55% of the games

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A plumber's daily earnings have a mean of $145 per day with a standard deviation of $16.50.
avanturin [10]

Step-by-step explanation:

The plumber's daily earnings have a mean of $145 per day with a standard deviation of

$16.50.

We want to find the probability that the plumber earns between $135 and

$175 on a given day, if the daily earnings follow a normal distribution.

That is we want to find P(135 <X<175).

Let us convert to z-scores using

z =  \frac{x -  \mu}{ \sigma}

This means that:

P(135  \: <  \: X  \: <  \: 175) = P( \frac{135 - 145}{16.5}  \: <  \:  z \:  <  \frac{175 - 145}{ 16.5} )

We simplify to get:

P(135  \: <  \: X  \: <  \: 175) = P(  - 0.61\: <  \:  z \:  <  1.82 )

From the standard n normal distribution table,

P(z<1.82)=0.9656

P(z<-0.61)=0.2709

To find the area between the two z-scores, we subtract to obtain:

P(-0.61<z<1.82)=0.9656-0.2709=0.6947

This means that:

P(135  \: <  \: X  \: <  \: 175) =0.69

The correct choice is C.

7 0
3 years ago
Question in attachment​
valkas [14]

Answer:

Hey Dude....

Step-by-step explanation:

This is ur answer.....

<h3><em>(a) Six</em></h3><h3><em>(a) Six(b) 120</em></h3><h3><em>(a) Six(b) 120(c) 720</em></h3>

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7 0
2 years ago
5) In a certain supermarket, a sample of 60 customers who used a self-service checkout lane averaged 5.2 minutes of checkout tim
Ne4ueva [31]

Answer:

\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643

S_p=2.940

t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751

df=60+72-2=130

p_v =P(t_{130}

Assuming a significance level of \alpha=0.05 we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

Step-by-step explanation:

Data given

Our notation on this case :

n_1 =60 represent the sample size for people who used a self service

n_2 =72 represent the sample size for people who used a cashier

\bar X_1 =5.2 represent the sample mean for people who used a self service

\bar X_2 =6.1 represent the sample mean people who used a cashier

s_1=3.1 represent the sample standard deviation for people who used a self service

s_2=2.8 represent the sample standard deviation for people who used a cashier

Assumptions

When we have two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2

The statistic is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

And t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

System of hypothesis

Null hypothesis: \mu_1 \geq \mu_2

Alternative hypothesis: \mu_1 < \mu_2

This system is equivalent to:

Null hypothesis: \mu_1 - \mu_2 \geq 0

Alternative hypothesis: \mu_1 -\mu_2 < 0

We can find the pooled variance:

\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643

And the deviation would be just the square root of the variance:

S_p=2.940

The statistic is given by:

t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751

The degrees of freedom are given by:

df=60+72-2=130

And now we can calculate the p value with:

p_v =P(t_{130}

Assuming a significance level of \alpha=0.05 we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

5 0
3 years ago
Emma decides to invest $990,000 in a period annuity that earns a 2.2% APR,
Pani-rosa [81]

Answer: B) 6462.32

Step-by-step explanation:

Compound interest can be calculated via the attachment.

Hope it helps <3

4 0
3 years ago
Read 2 more answers
A triangle has sides with lengths of 3 feet, 3 feet, and 5 feet. Is it a right triangle?
Anestetic [448]

Step-by-step explanation:

3^2 + 3^2 =5^2

9 +9 =25

18 =25

False 18 is not equal to 25.

7 0
2 years ago
Read 2 more answers
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