Arc CD is One-fourth of the circumference of a circle. What is the radian measure of the central angle?

(6.2N-8.3) +(9.1 + 1.4N)

lakkis [162]

Answer:

the answer is =7.6n+0.8

Step-by-step explanation:

hoped I helped

3 0

3 years ago

Solve the inequality -1/3h ≥ 7 A. h ≥ –21 B. h ≤ 7 1/3 C. h ≤ –21 D. h ≥ -2 1/3

JulsSmile [24]

The Answer Is C Fam...................

8 0

3 years ago

Read 2 more answers

Which of the following graph is the solution -10<3x-4<8

satela [25.4K]

Answer:

( − 2 , 4 )

Step-by-step explanation:

There were no graphs listed in your question. So...

Convert the inequality to interval notation.

6 0

3 years ago

⦁ In a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football. Construct a 95% confiden

fomenos

Answer:

95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

Step-by-step explanation:

We are given that in a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football.

Firstly, the pivotal quantity for 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = proportion of adults in the United States whose favorite sport to watch is football in a sample of 1219 adults = $\frac{354}{1219}$

n = sample of US adults = 1291

p = population proportion of adults

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

significance level are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ]

= [ $\frac{354}{1219}-1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} }$ , $\frac{354}{1219}+1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} }$ ]

= [0.265 , 0.316]

Therefore, 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

5 0

2 years ago

Cards are drawn, one at a time, from a standard deck; each card is replaced before the next one is drawn. Let X be the number of

steposvetlana [31]

Cards are drawn, one at a time, from a standard deck; each card is replaced before the next one is drawn. Let X be the number of draws necessary to get an ace. Find E(X) is given in the following way

Step-by-step explanation:

From a standard deck of cards, one card is drawn. What is the probability that the card is black and a jack? P(Black and Jack) P(Black) = 26/52 or ½ , P(Jack) is 4/52 or 1/13 so P(Black and Jack) = ½ * 1/13 = 1/26

A standard deck of cards is shuffled and one card is drawn. Find the probability that the card is a queen or an ace.

P(Q or A) = P(Q) = 4/52 or 1/13 + P(A) = 4/52 or 1/13 = 1/13 + 1/13 = 2/13

WITHOUT REPLACEMENT: If you draw two cards from the deck without replacement, what is the probability that they will both be aces?

P(AA) = (4/52)(3/51) = 1/221.

WITHOUT REPLACEMENT: What is the probability that the second card will be an ace if the first card is a king?

P(A|K) = 4/51 since there are four aces in the deck but only 51 cards left after the king has been removed.

WITH REPLACEMENT: Find the probability of drawing three queens in a row, with replacement. We pick a card, write down what it is, then put it back in the deck and draw again. To find the P(QQQ), we find the

probability of drawing the first queen which is 4/52.

The probability of drawing the second queen is also 4/52 and the third is 4/52.

We multiply these three individual probabilities together to get P(QQQ) =

P(Q)P(Q)P(Q) = (4/52)(4/52)(4/52) = .00004 which is very small but not impossible.

Probability of getting a royal flush = P(10 and Jack and Queen and King and Ace of the same suit)

5 0

2 years ago