Ron wants to buy a video game with a selling price of $48. It is on sale for 50% off. The sales tax in his state is 4.5%.

nika2105 [10]

Answer:

The answer is C..........

4 0

3 years ago

The ordered pairs (1, 1), (2, 8), (3, 27), (4, 64), and (5, 125) represent a function. What is a rule that represents this funct

Zarrin [17]

The functions would be:

D. y=x³

We can check it out.

(1,1); x=1; y=1 ⇒1=(1)³=1*1*1=1
(2,8); x=2; y=8 ⇒8=(2)³=2*2*2=8
(3,27); x=3; y=27 ⇒27=3³=3*3*3=27
(4,64): x=4; y=64 ⇒64=4³=4*4*4=64
(5,125); x=5; y=125 ⇔ 125=5³=5*5*5=125

8 0

3 years ago

Mathew basketball team won 68% of the games they played if they won 17 games how many did they play

STALIN [3.7K]

Answer:

25

Step-by-step explanation:

turn 68% into a decimal (or keep as a percentage, your choice) and divide

17 divided by 0.68 (68% as a decimal) = 25

6 0

2 years ago

During the 7th examination of the Offspring cohort in the Framingham Heart Study, there were 1219 participants being treated for

AlexFokin [52]

Answer:

95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

Step-by-step explanation:

We are given that during the 7th examination of the Offspring cohort in the Framing ham Heart Study, there were 1219 participants being treated for hypertension and 2,313 who were not on treatment.

The sample proportion is : $\hat p$ = x/n = 1219/3532 = 0.345

Firstly, the pivotal quantity for 95% confidence interval for the proportion of the population is given by;

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

where, $\hat p$ = sample proportion = 0.345

n = sample of participants = 3532

p = population proportion

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% level of

significance 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= [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.345-1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } }$ , $0.345+1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } }$ ]

= [0.3293 , 0.3607]

Hence, 95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

6 0

3 years ago

Tim has to write two papers worth 50 points each. To pass, he has to get at least 45% of the total points. However, scoring less

lions [1.4K]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Based on the above situation the graph that represents the region in which Tim fails is the graph A.

4 0

3 years ago

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