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2 years ago

5

For a certain instant lottery game comma the odds in favor of a win instant lottery game, the odds in favor of a win are given a

s 59 to 41. Express the indicated degree of likelihood as a probability value between 0 and 1 inclusive. The probability is

2 answers:

Fittoniya [83]2 years ago

6 0

Answer:

0.41

Step-by-step explanation:

p = 41/100 = 0.41

const2013 [10]2 years ago

4 0

Answer:

0.59

Step-by-step explanation:

The odds in favor of winning the lottery game are 59 to 41. This means for every 59 chances of winning the lottery game there are 41 chances of losing the game.

We have to find the probability of winning the lottery game. Since there are 59 chances to win the game and 41 chances to loss the game, the total number of games come out to be 59 + 41 = 100

i.e.

Out of every 100 games, there is a chance of winning 59 games and losing 41 games

Therefore,

The probability of winning the lottery game would be = 59/100 = 0.59

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I WILL GAVE BRAINLIEST

Arte-miy333 [17]

Answer:

x < -14

Step-by-step explanation:

Step 1: Subtract 6 from both sides.

$-3x+6-6 > 48-6$
$-3x > 42$

Step 2: Divide both sides by -3 and flip the inequality sign.

$-3x/-3 > 42/-3$
$x < -14$

Therefore, the answer is x < -14.

4 0

1 year ago

Read 2 more answers

For each question, each pair of triangles are similar but not drawn to scale. Calculate any lettered lengths.

Karolina [17]

Answer:

a. 20

b. 5

Step-by-step explanation:

The scale factor is 5 because 6×5=30

a. 4×5=20

b. 25÷5=5

8 0

2 years ago

The following results come from two independent random samples taken of two populations.

photoshop1234 [79]

Answer:

$(a)\ \bar x_1 - \bar x_2 = 2.0$

$(b)\ CI =(1.0542,2.9458)$

$(c)\ CI = (0.8730,2.1270)$

Step-by-step explanation:

Given

$n_1 = 60$ $n_2 = 35$

$\bar x_1 = 13.6$ $\bar x_2 = 11.6$

$\sigma_1 = 2.1$ $\sigma_2 = 3$

Solving (a): Point estimate of difference of mean

This is calculated as: $\bar x_1 - \bar x_2$

$\bar x_1 - \bar x_2 = 13.6 - 11.6$

$\bar x_1 - \bar x_2 = 2.0$

Solving (b): 90% confidence interval

We have:

$c = 90\%$

$c = 0.90$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.90$

$\alpha = 0.10$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.10/2}$

$z_{\alpha/2} = z_{0.05}$

The z score is:

$z_{\alpha/2} = z_{0.05} =1.645$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.645 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.645 * \sqrt{0.3306}$

$2.0 \± 0.9458$

Split

$(2.0 - 0.9458) \to (2.0 + 0.9458)$

$(1.0542) \to (2.9458)$

Hence, the 90% confidence interval is:

$CI =(1.0542,2.9458)$

Solving (c): 95% confidence interval

We have:

$c = 95\%$

$c = 0.95$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.95$

$\alpha = 0.05$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.05/2}$

$z_{\alpha/2} = z_{0.025}$

The z score is:

$z_{\alpha/2} = z_{0.025} =1.96$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.96 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.96* \sqrt{0.3306}$

$2.0 \± 1.1270$

Split

$(2.0 - 1.1270) \to (2.0 + 1.1270)$

$(0.8730) \to (2.1270)$

Hence, the 95% confidence interval is:

$CI = (0.8730,2.1270)$

8 0

3 years ago

One more guys. Please make sure you use PEMDAS!

Alexxx [7]

Answer:

-14

Step-by-step explanation:

Replace the 2 for each k.

$\frac{1}{4}2^{5} - 3k^{3} + k^{2} - k$

then solve it using PEDMAS: exponents first

= -14

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2 years ago

How to do a diagram in math for kelly correctly answered 80% of the 20 questions on the test

Dvinal [7]

Step-by-step explanation:

80:100=x:20

x=80×20/100=16

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2 years ago