Help please!

Two runners are saving money to attend a marathon. The first runner has $112 in savings, received a $45 gift from a friend, and

Volgvan

Answer:

The correct answer is the last option, that is,

112 + 25m + 45 = 50 + 60m

Step-by-step explanation:

We have been given that the first runner has $112 in savings, received a $45 gift from a friend, and will save $25 each month. Therefore, amount of money in the account of first running after m months will be: 112 + 45 + 25m

We have been given that the second runner has $50 in savings and will save $60 each month. Therefore, amount of money in the account of second running after m months will be: 50 + 60m

In order for amount of money to be equal in accounts of both the runners, we set up: 112 + 45 + 25m = 50 + 60m

Upon rewriting the left hand side using commutative law, we get:

112 + 25m + 45 = 50 + 60m

Therefore, we can see that the last option is the correct answer.

I hope this helped. I am sorry if you get this wrong.

4 0

3 years ago

Read 2 more answers

Joe's lunch at a restaurant costs $13.00 without tax. He leaves the waiter a tip of 17% of the cost of the lunch, without tax. W

choli [55]

Answer: $15.21

Step-by-step explanation:

Given

Joe's lunch cost $13.00 without tax

He tipped the waiter with 17% of bill without tax

So, the amount of tip is

$\Rightarrow 13\times 17\%\\\Rightarrow 13\times 0.17\\\Rightarrow \$2.21$

Total amount that food costs without tax is

$\Rightarrow 13+2.21\\\Rightarrow \$15.21$

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

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