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3 years ago

5

Genevieve loves 30% off wing night. She orders 12 wings that normally cost 1.25$ each. If the sales tax rate is 6.75%, how much

will she pay in taxes?

2 answers:

Dmitry_Shevchenko [17]3 years ago

8 0

Approximately 71 cents

pickupchik [31]3 years ago

6 0

Aproxementaley 71 cents

You might be interested in

(a) How many integers from 1 through 1,000 are multiples of 2 or multiples of 9?

DENIUS [597]

Answer:

(a)There are 500 integers from 1 through 1,000 are multiples of 2.

There are 111 integers from 1 through 1,000 are multiples of 9.

(b)0.611

Step-by-step explanation:

Given the set of Integers from 1 through 1000

The least multiple of 2 is 2 and the highest multiple of 2 in the interval is 1000.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 2,4,6,... is an Arithmetic Progression,

where first term, a =2, common difference, d =2

Nth term of an A.P,

$U_n= a+(n-1)d\\1000=2+2(n-1)\\1000=2+2n-2\\1000=2n\\n=500$

There are 500 integers from 1 through 1,000 are multiples of 2.

Similarly,

Given the set of Integers from 1 through 1000

The least multiple of 9 is 9 and the highest multiple of 9 in the interval is 999.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 9,18,27,... is an Arithmetic Progression,

where first term, a =9, common difference, d =9

Nth term of an A.P,

$U_n= a+(n-1)d\\999=9+9(n-1)\\999=9+9n-9\\999=9n\\n=111$

There are 111 integers from 1 through 1,000 are multiples of 9.

(b)Probability that an integer selected at random is a multiple of 2 or a multiple of 9.

There are a total of 1000 numbers between from 1 through 1,000

n(S)=1000

n(Multiples of 2)=500

n(Multiples of 9)=111

Therefore:

$P(\text{Multiples of 9}) \:OR\: P(\text{Multiples of 2})=\frac{111}{1000} + \frac{500}{1000} \\=\frac{611}{1000} \\=0.611$

7 0

3 years ago

A one-parameter family of solutions of the de p' = p(1 − p) is given below.

tensa zangetsu [6.8K]

Answer:

A solution curve pass through the point (0,4) when $c_{1} = -\frac{4}{3}$ .

There is not a solution curve passing through the point(0,1).

Step-by-step explanation:

We have the following solution:

$P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}$

Does any solution curve pass through the point (0, 4)?

We have to see if P = 4 when t = 0.

$P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}$

$4 = \frac{c_{1}}{1 + c_{1}}$

$4 + 4c_{1} = c_{1}$

$c_{1} = -\frac{4}{3}$

A solution curve pass through the point (0,4) when $c_{1} = -\frac{4}{3}$ .

Through the point (0, 1)?

Same thing as above

$P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}$

$1 = \frac{c_{1}}{1 + c_{1}}$

$1 + c_{1} = c_{1}$

$0c_{1} = 1$

No solution.

So there is not a solution curve passing through the point(0,1).

3 0

3 years ago

What is 6/15 written as a decimal?

Allushta [10]

It will be C cause you divid 15 and 6

3 0

3 years ago

Read 2 more answers

) One year, professional sports players salaries averaged $1.5 million with a standard deviation of $0.9 million. Suppose a samp

Nutka1998 [239]

Answer:

Probability that the average salary of the 400 players exceeded $1.1 million is 0.99999.

Step-by-step explanation:

We are given that one year, professional sports players salaries averaged $1.5 million with a standard deviation of $0.9 million.

Suppose a sample of 400 major league players was taken.

<em>Let </em> $\bar X$ <em> = sample average salary</em>

The z-score probability distribution for sample mean is given by;

Z = $\frac{ \bar X -\mu}{{\frac{\sigma}{\sqrt{n} } }} }$ ~ N(0,1)

where, $\mu$ = mean salary = $1.5 million

$\sigma$ = standard deviation = $0.9 million

n = sample of players = 400

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, probability that the average salary of the 400 players exceeded $1.1 million is given by = P( $\bar X$ > $1.1 million)

P( $\bar X$ > $1.1 million) = P( $\frac{ \bar X -\mu}{{\frac{\sigma}{\sqrt{n} } }} }$ > $\frac{ 1.1-1.5}{{\frac{0.9}{\sqrt{400} } }} }$ ) = P(Z > -8.89) = P(Z < 8.89)

<em>Now, in the z table the maximum value of which probability area is given is for critical value of x = 4.40 as 0.99999. So, we can assume that the above probability also has an area of 0.99999 or nearly close to 1.</em>

Therefore, probability that the average salary of the 400 players exceeded $1.1 million is 0.99999.

4 0

2 years ago

What is the ratio of the number of days in the week that do not begin with the letter "T" to the number of days in the week that

serg [7]

Answer:

5:7

Step-by-step explanation:

The days in the week whose names don't begin with 'T' are Sunday, Monday, Wednesday, Friday, and Saturday. I counted them, and found that there are five (5) of them. Then I counted the number of days in a full week, no matter how they're spelled. I counted seven (7) that time. The ratio of 5 to 7 is. 5:7 or 5/7.

8 0

2 years ago