Hamburger is sold 2 pounds for $4. If Alicia buys 10 pounds of hamburgers, how much will she pay?

Vinil7 [7]

Note: The third term of the sequence should be $\dfrac{27}{2}$ instead of $\dfrac{37}{2}$ , otherwise the sequence has no common ratio.

Given:

The given sequence is

$24,18,\dfrac{27}{2},\dfrac{81}{8}$

To find:

The common ratio of the given sequence.

Solution:

The quotient of each pair of consecutive terms are:

$\dfrac{18}{24}=\dfrac{3}{4}$

Similarly,

$\dfrac{\dfrac{27}{2}}{18}=\dfrac{27}{36}$

$\dfrac{\dfrac{27}{2}}{18}=\dfrac{3}{4}$

And,

$\dfrac{\dfrac{81}{8}}{\dfrac{27}{2}}=\dfrac{81}{8}\times \dfrac{2}{27}$

$\dfrac{\dfrac{81}{8}}{\dfrac{27}{2}}=\dfrac{3}{4}$

Therefore, the common ratio of the given sequence is $\dfrac{3}{4}$ or 0.75.

8 0

3 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

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

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

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

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

= [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

4) Choose the correct inequality for this situation.

Ksenya-84 [330]

Answer:

B. 21x + 40 ≤ 124

Step-by-step Explanation:

Maximum amount budgeted = $124 (this means they can't spend more than this)

x = number of people

Cost per head = $21

Given that Mr Walter already spent $40, which is part of the money budgeted, the number of people that can go canoeing cam be expressed with the following inequality:

21x + 40 ≤ 124

(note: the amount total to be spent will either be equal to or greater than $124, because it's the maximum amount budgeted for spending).

6 0

3 years ago

WHATS 9+10

algol [13]

Answer:

19

Step-by-step explanation:

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21

3 0

3 years ago

Read 2 more answers

Your bedroom wall is 171 inches wide. You want to hang two posters, each 42 inches across, with an equal space between each post

Readme [11.4K]

Your answer should be 29

3 0

3 years ago