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wel
4 years ago
10

The standard tip in a restaurant is 15% of the bill before tax. Many people in California find the tip by doubling the sales tax

, which is 8.25%. By how many cents does this exceed the standard tip on a bill of 60% before tax?
Mathematics
1 answer:
cricket20 [7]4 years ago
3 0
To answer this you could do a couple things. 1. Actually calculate both tips and then subtract them. Here is the math ( assuming it is $60, not 60% that you mean). a. 0.15 x 60 = $9. b. 0.0825 x 2 x 60=$9.90. c. 9.90-9.00= $0.90 difference.

The other strategy would be to find the difference on the percents and then multiply by $60. a. 8.25%x2=16.5%-15%. Difference of 1.5%. b. 0.015 x $60=$0.90.
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Help please (will chose brainliest).
lara [203]
Answers with Explanation.


i. If we raise a number to an exponent of 1, we get the same number.

{10}^{1}  = 10

ii. If we raise 10 to an exponent of 2, it means we multiply 10 by itself two times.

{10}^{2}  = 10 \times 10 = 100

iii. If we raise 10 to an exponent of 3, it means we multiply 10 by itself three times.



{10}^{3}  = 10 \times  {10}^{2}  = 10 \times 100 = 1000

iv. If we raise 10 to an exponent of 4, it means we multiply 10 by itself four times.


{10}^{4}  =  10 \times {10}^{3} = 10 \times 1000 =  10000

v. If we raise 10 to an exponent of 5, it means we multiply 10 by itself five times.



{10}^{5}  = 10 \times  {10}^{4}  = 10 \times 10000 = 100000

{10}^{5}  = 100000

vi. Recall that,

{a}^{ - m}  =  \frac{1}{ {a}^{m} }
We apply this law of exponents to obtain,

{10}^{ - 2}  =  \frac{1}{{10}^{2} }  =  \frac{1}{100}

vii. We apply
{a}^{ - m}  =  \frac{1}{ {a}^{m} }
again to obtain,


{10}^{ - 3}  =  \frac{1}{ {10}^{3} }  =  \frac{1}{1000}





5 0
3 years ago
Pls help me or me fail
Mrac [35]

Answer:

Cumin has the least price per ounce

Step-by-step explanation:

1. Divide each spice price to the size

Cumin -> 4.80/1.5= 3.2  

Ginger -> 3.60/ 0.8= 4.5  

Nutmeg -> 6.30/ 1.8= 3.5  

Thyme -> 4.50/ 1.25= 3.6  

5 0
4 years ago
What are the dimensions of the product?<br> a 1 x 2<br> b 2 x 2<br> c 3 x 2<br> d 3 x 3
jenyasd209 [6]
21 of your taking about area
3 0
3 years ago
Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie
ludmilkaskok [199]

Answer:

\lambda \geq 6.63835

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that X \sim Poisson(\lambda)

The probability mass function for the random variable is given by:

f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...

And f(x)=0 for other case.

For this distribution the expected value is the same parameter \lambda

E(X)=\mu =\lambda

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

P(X\geq 2)=1-P(X

Using the pmf we can find the individual probabilities like this:

P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}

P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}

And replacing we have this:

P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]

P(X\geq 2)=1-e^{-\lambda}(1+\lambda)

And we want this probability that at least of 99%, so we can set upt the following inequality:

P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99

And now we can solve for \lambda

0.01 \geq e^{-\lambda}(1+\lambda)

Applying natural log on both sides we have:

ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)

ln(0.01) \geq -\lambda+ln(1+\lambda)

\lambda-ln(1+\lambda)+ln(0.01) \geq 0

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}

Where :

f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)

f'(x_n)=1-\frac{1}{1+\lambda}

Iterating as shown on the figure attached we find a final solution given by:

\lambda \geq 6.63835

4 0
3 years ago
Given that M is the midpoint of L and P. If LP=4x-1, LM=8, find x and LP
Alenkinab [10]

Answer:

LP = 16 units (8 + 8)

x = 4.25 units

Step-by-step explanation:

8 0
3 years ago
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