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2 years ago

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JACKIE COLLECTS DATA THROUGTH THE BASEBALL SEASON AND USED THESE DATA TO DETERMINE THE PROBABILITIES OF DIFFRENT TEAMS PLAYING I

N THE LEAGUE CHAMPIONSHIP GAME,THE PROBABILITTIES FOR HER FAVORITE TEAMS PLAYING IN THE CHAMPIONSHIP GAME ARE SHOWN BELOW WHAT TEAMS IS LIKELY TO WIN

1 answer:

Citrus2011 [14]2 years ago

4 0

In this question, we want to compare several numbers with different denominators and find out which number is the least. To compare this number, we have to change the denominator into the same number by finding the least common multiple (LCM) of the 4 numbers. The factor of each number will be:

3= 3 ^1

5= 5^1

8= 2 * 2 * 2 = 2^3

2= 2^1

We can find the LCM by multiplying a higher exponent of each prime number. The LCM will be:3^1 * 5^1 * 2^3 = 120

Each number will be:

Tiger= 2/3 * 40/40= 80/120

Redbird = 4/5 * 24/24= 96/120

Bulldogs = 3/8 * 15/15 = 45/120

Titans = 1/2 * 60/60 = 60/120

As you can see, the team with the lowest chance to play is Bulldogs = 45/120

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The amount of money spent on textbooks per year for students is approximately normal.

Ostrovityanka [42]

Answer:a

a

$336.04 < \mu < 443.96$

b

The margin of error will increase

c

The margin of error will decreases

d

The 99% confidence interval is $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are told that

The sample size $n = 19$

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

So

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> $E = 53.96$

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

$n \ \ \alpha \ \ \frac{1}{E^2 }$

So if the sample size increases the margin of error decrease

The sample proportion is mathematically represented as

$\r p = \frac{210}{500}$

$\r p = 0.42$

Given that the confidence level is 0.99 the level of significance is $\alpha = 0.01$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} }* \sqrt{ \frac{\r p (1- \r p )}{n} }$

=> $E = 0.42 * \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }$

=> $E = 0.0093$

The 99% confidence interval is

$\r p - E < p < \r p + E$

$0.42 - 0.0093 < p < 0.42 + 0.0093$

$0.4107 < p < 0.4293$

4 0

3 years ago

four friends bought 10 lbs of nuts and shared them equally. How many pounds if nuts did each friend get

Marianna [84]

You need to do 10 divided by 4. Each friend would get 2.5 pounds of a nut.

5 0

3 years ago

Please help ASAP 50 points :(((

Katyanochek1 [597]

Answer:

(f o g) (x) = 36x² + 3

(g o f) (x) = 6x² + 18

Step-by-step explanation:

f(x) is x² + 3

g(x) is 6x

(f o g) (x) means for all xs in f replace it with g(x) or 6x

(f o g) (x) = (6x)² + 3 - see 6x is in place of x and 6x is g(x)

(f o g) (x) = 36x² + 3

(g o f) (x) = 6(x² + 3)

(g o f) (x) = 6x² + 18

8 0

3 years ago

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If f(x)=3^x+10 and g(x)=2x-4, find (f+g)(x)

Semmy [17]

Answer:

4

Step-by-step explanation:

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2 years ago

Find the slope of the following graph and write ur results in the empty box.

Nataly [62]

Slope is 2 because it goes up by two and over by one

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2 years ago