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3 years ago

12

The Bears have won 7 and tied 2 of their last 13 games. They have not forfeited any games. Which ratio correctly compares their

wins to losses?

1 answer:

nataly862011 [7]3 years ago

8 0

Answer:

7:4

Step-by-step explanation:

7:4 because if 13 is the total amount of games, that minus the amount of games won and tied equals the amount of games loss which is 4. Since it says wins to losses, the wins will come first in the ratio. Therefore the correct answer would be 7:4.

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Plzzz help don’t understand plz explain

kherson [118]

So the easiest way to do this problem is to put your equation in linear form, meaning y = the rest of the equation.

I'm going to show the work to do that below:

2x - y = 4

-2x -2x (Here I am subtracting 2x)

- y = 4 - 2x

(-1)(- y) = (-1)(4 - 2x) (Now I'm multiplying both sides by -1 to make y positive)

y = 2x - 4 (And just for neat purposes I'm going to put the 2x in front of the 4)

Now you have your equation ( y=2x-4 ) and you're ready to put it on the points of your graph.

Start on the point (0, -4) to represent the y axis.

y = 2x - 4

Then, your rise (2) over your run (1)

so your coordinates for a line will be:

(0, -4) (-2, 1) (1, 0)

You only need to plot these three to connect your line.

I appologize for my lack of visuals i understand this might be confusing, however if you have any further questions feel free to let me know!

8 0

4 years ago

6) Supplementary Exercise 5.51

tresset_1 [31]

Answer:

$P(X \le 4) = 0.7373$

$P(x \le 15) = 0.0173$

$P(x > 20) = 0.4207$

$P(20\ge x \le 24)= 0.6129$

$P(x = 24) = 0.0236$

$P(x = 15) = 1.18\%$

Step-by-step explanation:

Given

$p = 80\% = 0.8$

The question illustrates binomial distribution and will be solved using:

$P(X = x) = ^nC_xp^x(1 - p)^{n-x}$

Solving (a):

Given

$n =5$

Required

$P(X\ge 4)$

This is calculated using

$P(X \le 4) = P(x = 4) +P(x=5)$

This gives:

$P(X \le 4) = ^5C_4 * (0.8)^4*(1 - 0.8)^{5-4} + ^5C_5*0.8^5*(1 - 0.8)^{5-5}$

$P(X \le 4) = 5 * (0.8)^4*(0.2)^1 + 1*0.8^5*(0.2)^0$

$P(X \le 4) = 0.4096 + 0.32768$

$P(X \le 4) = 0.73728$

$P(X \le 4) = 0.7373$ --- approximated

Solving (b):

Given

$n =25$

i)

Required

$P(X\le 15)$

This is calculated as:

$P(X\le 15) = 1 - P(x>15)$ --- Complement rule

$P(x>15) = P(x=16) + P(x=17) + P(x =18) + P(x = 19) + P(x = 20) + P(x = 21) + P(x = 22) + P(x = 23) + P(x = 24) + P(x = 25)$

$P(x > 15) = {25}^C_{16} * p^{16}*(1-p)^{25-16} +{25}^C_{17} * p^{17}*(1-p)^{25-17} +{25}^C_{18} * p^{18}*(1-p)^{25-18} +{25}^C_{19} * p^{19}*(1-p)^{25-19} +{25}^C_{20} * p^{20}*(1-p)^{25-20} +{25}^C_{21} * p^{21}*(1-p)^{25-21} +{25}^C_{22} * p^{22}*(1-p)^{25-22} +{25}^C_{23} * p^{23}*(1-p)^{25-23} +{25}^C_{24} * p^{24}*(1-p)^{25-24} +{25}^C_{25} * p^{25}*(1-p)^{25-25}$

$P(x > 15) = 2042975 * 0.8^{16}*0.2^9 +1081575* 0.8^{17}*0.2^8 +480700 * 0.8^{18}*0.2^7 +177100 * 0.8^{19}*0.2^6 +53130 * 0.8^{20}*0.2^5 +12650 * 0.8^{21}*0.2^4 +2300 * 0.8^{22}*0.2^3 +300 * 0.8^{23}*0.2^2 +25* 0.8^{24}*0.2^1 +1 * 0.8^{25}*0.2^0$

$P(x > 15) = 0.98266813045$

So:

$P(X\le 15) = 1 - P(x>15)$

$P(x \le 15) = 1 - 0.98266813045$

$P(x \le 15) = 0.01733186955$

$P(x \le 15) = 0.0173$

ii)

$P(x>20)$

This is calculated as:

$P(x>20) = P(x = 21) + P(x = 22) + P(x = 23) + P(x = 24) + P(x = 25)$

$P(x > 20) = 12650 * 0.8^{21}*0.2^4 +2300 * 0.8^{22}*0.2^3 +300 * 0.8^{23}*0.2^2 +25* 0.8^{24}*0.2^1 +1 * 0.8^{25}*0.2^0$

$P(x > 20) = 0.42067430925$

$P(x > 20) = 0.4207$

iii)

$P(20\ge x \le 24)$

This is calculated as:

$P(20\ge x \le 24) = P(x = 20) + P(x = 21) + P(x = 22) + P(x =23) + P(x = 24)$

$P(20\ge x \le 24)= 53130 * 0.8^{20}*0.2^5 +12650 * 0.8^{21}*0.2^4 +2300 * 0.8^{22}*0.2^3 +300 * 0.8^{23}*0.2^2 +25* 0.8^{24}*0.2^1$

$P(20\ge x \le 24)= 0.61291151859$

$P(20\ge x \le 24)= 0.6129$

iv)

$P(x = 24)$

This is calculated as:

$P(x = 24) = 25* 0.8^{24}*0.2^1$

$P(x = 24) = 0.0236$

Solving (c):

$P(x = 15)$

This is calculated as:

$P(x = 15) = {25}^C_{15} * 0.8^{15} * 0.2^{10}$

$P(x = 15) = 3268760 * 0.8^{15} * 0.2^{10}$

$P(x = 15) = 0.01177694905$

$P(x = 15) = 0.0118$

Express as percentage

$P(x = 15) = 1.18\%$

The calculated probability (1.18%) is way less than the advocate's claim.

Hence, we do not believe the claim.

5 0

3 years ago

8 1/3% of 72 <br> 10 points please help ASAP

olasank [31]

Answer:

6

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

In a particular list of three-digit perfect squares, the first perfect square can be turned into each of the others by rearrangi

kiruha [24]

The Answer is Three.

That's the largest nmber of 3-digit perfect squares that could be on the list.

The list is so;

169 = 13²

196 = 14²

961 = 31²

The thre numbers, 1, 6, and 9 can be rearranged three ways to form three 3-digit perfect squares in 169, 196, and 961. No other arrangement of a 3-digit perfect square can yield more or equal.

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3 years ago

9 to the x power=3 to the 40th power<br><br> Someone explain how??

Mrrafil [7]

9^x = 3^2x

so...

2x = 40

x = 20

8 0

3 years ago