How many fourteenths are in 2/7

Ben spent 1/6 of his money on a burger, fries, and a drink. Then he spent half of the money he had left: $5 on a magazine, $8.25

Ket [755]

The amount of money Ben had to begin with after spending 1/6 and 1/2 of it is 57 dollars.

<h3>How to find the how much money he had with an equation?</h3>

let

x = amount he had to begin with

He spent 1/6 of his money on a burger, fries, and a drink. Therefore,

amount spent on burger, fries, and a drink = 1 / 6 x

Hence,

amount he had left = x - 1 / 6 x =6x - x /6 = 5 / 6 x

Then he spent half of the money he had left.

1 / 2(5 /6 x) = 5 + 8.25 + 10.50

5 / 12 x = 23.75

cross multiply

5x = 23.75 × 12

5x = 285

divide both sides by 5

x = 285 / 5

x = 57

Therefore, the amount of money he have to begin with is $57.

learn more on equation here: brainly.com/question/5718696

4 0

2 years ago

A card game has 50 cards. After dealing 7 cards to each player, Tupi has 15 cards left over. Solve the equation 50 7p = 15 to fi

mafiozo [28]

Answer:

50 - 7p = 15....subtract 50 from both sides

-7p = 15 - 50

-7p = - 35...divide both sides by -7

p = -35/-7

p = 5 <=== there are 5 players

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

I’m stuck on this and I need help...

Nat2105 [25]

Answer:

D

Step-by-step explanation:

The ratio of corresponding sides are equal, that is

$\frac{4}{8}$ = $\frac{y}{14}$

3 0

4 years ago

1. In order to get more female customers, a new clothing store offers free gourmet coffee and pastry to its customers. The avera

ioda

Answer:

No, the manager is not correct based on the 95% confidence interval.

Step-by-step explanation:

We are given that the average daily revenue over the past five-week period has been $1,080 with a standard deviation of $260, i.e.; X bar = $1080 and s = $260 and sample size, n = 35 .

The Pivotal quantity for 95% confidence interval is given by;

$\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, X bar = sample mean = $1080

s = sample standard deviation = $260

n = sample size = 35 {five-week}

So, 95% confidence interval for average daily revenue, $\mu$ is given by;

P(-2.032 < $t_3_4$ < 2.032) = 0.95

P(-2.032 < $\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }$ < 2.032) = 0.95

P(-2.032 * ${\frac{s}{\sqrt{n} }$ < ${Xbar - \mu}$ < 2.032 * ${\frac{s}{\sqrt{n} }$ ) = 0.95

P(X bar - 2.032 * ${\frac{s}{\sqrt{n} }$ < $\mu$ < X bar + 2.032 * ${\frac{s}{\sqrt{n} }$ ) = 0.95

95% confidence interval for $\mu$ = [ X bar - 2.032 * ${\frac{s}{\sqrt{n} }$ , X bar + 2.032 * ${\frac{s}{\sqrt{n} }$ ]

= [ 1080 - 2.032 * ${\frac{260}{\sqrt{35} }$ , 1080 + 2.032 * ${\frac{260}{\sqrt{35} }$ ]

= [ 990.70 , 1169.30 ]

<em>No, the manager is not correct based on the fact that the coffee and pastry strategy would lead to an average daily revenue of $1,200 because the calculate 95% confidence interval does not include value of $1200.</em>

Therefore, the store manager believe is not correct.

8 0

4 years ago

Which expression is equal to

Burka [1]

Answer:

4^5/ 4^-9

Step-by-step explanation:

the third one

3 0

3 years ago

Read 2 more answers