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2 years ago

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John and 3 friends are going out for pizza. They split one pizza and 4 large drinks. The pizza cost $11.50. They spend a total o

f 16.90. Find the cost of the one large drink.

2 answers:

erma4kov [3.2K]2 years ago

7 0

What we do first is subtract 11.50 from 16.90.

16.90 - 11.50 = 5.4.

<h3>$5.40 is how much the four drinks cost in total.</h3>

If you wanna find the amount for each drink separately, divide 5.4 by 4.

5.4 / 4 = 1.35.

<h3>Each drink cost $1.35.</h3>

poizon [28]2 years ago

4 0

Answer: $1.68

Step-by-step explanation: First you subtract 16.90 by 11.50 which gives you $5.04 Next, you divide $5.04 by the amount of friends there are (3) Then that gives you the total of $1.68 total cost for drinks.

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A whole number n which when divided by 4 gives 3 as remainder. What will be the remainder when 2n is divided by 4?.

jeyben [28]

When 2n is divided by 4 the remainder will be 2

Division is a simple operation in which a number is divided.

Given,

A whole number n which when divided by 4 gives 3 as remainder

We know,

Dividend = (Divisor × quotient)+ Remainder

Here,

Dividend = n

Quotient = 4

Remainder =3

Consider the divisor as a

Then,

n = 4a+3

Multiply both side by 2

2n= 2(4a+3)

2n= 8a+6

Split the term 6 and take common outside

2n= 8a+4+2

2n= 4(2a+1)+2

So, when 2n is divided by 4

Divisor = 2a+1

Remainder = 2

Hence, when 2n is divided by 4 the remainder will be 2

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1 year ago

Interest centers around the life of an electronic component. Let A be the event that the component fails a particular test and B

nekit [7.7K]

Answer: a. 0.61

b. 0.37

c. 0.63

Step-by-step explanation:

From the question,

P(A) = 0.39 and P(B) = 0.24

P(success) + P( failure) = 1

A) What is the probability that the component does not fail the test?

Since A is the event that the component fails a particular test, the probability that the component does not fail the test will be P(success). This will be:

= 1 - P(A)

= 1 - 0.39

= 0.61

B) What is the probability that a component works perfectly well (i.e., neither displays strain nor fails the test)?

This will be the probability that the component does not fail the test minus the event that the component displays strain but does not actually fail. This will be:

= [1 - P(A)] - P(B)

= 0.61 - 0.24

= 0.37

C) What is the probability that the component either fails or shows strain in the test?

This will simply be:

= 1 - P(probability that a component works perfectly well)

= 1 - 0.37

= 0.63

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2 years ago

Use the conditional statement to answer the question.

anygoal [31]

The statement is true, but the inverse is false.

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3 years ago

What is the value of (7-12)(-8-22)

Fed [463]

Answer:

The Answer to this problem is 150

Step-by-step explanation:

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2 years ago

Read 2 more answers

F(x) = (128/127)(1/2)x, x = 1,2,3,...7. determine the requested values: round your answers to three decimal places (e.g. 98.765)

Marrrta [24]

A.
$\mathbb P(X\le 1)=\mathbb P(X=1)=\dfrac{128}{127}\left(\dfrac12\right)^1=\dfrac{64}{127}$

b.
$\mathbb P(X>1)=1-\mathbb P(X\le1)=1-\dfrac{64}{127}=\dfrac{63}{127}$

c.
$\mathbb E(X)=\displaystyle\sum_{x=1}^7 x\,f_X(x)=\frac{64}{127}\sum_{x=1}^7 x\left(\frac12\right)^{x-1}$

Suppose $f(y)=\displaystyle\sum_{x=0}^7 y^x$ . Then $f'(y)=\displaystyle\sum_{x=1}^7 xy^{x-1}$ . So if we can find a closed form for $f(y)$ , in terms of $y$ , we can find $\mathbb E(X)$ by evaluating the derivative of $f(y)$ at $y=\dfrac12$ .

$f(y)=\displaystyle\sum_{x=0}^7 y^x=y^0+y^1+y^2+\cdots+y^6+y^7$
$y\,f(y)=y^1+y^2+y^3+\cdots+y^7+y^8$
$f(y)-y\,f(y)=y^0-y^8$
$(1-y)f(y)=1-y^8$
$f(y)=\dfrac{1-y^8}{1-y}$
$\implies f'(y)=\dfrac{7y^8-8y^7+1}{(1-y)^2}$
$\implies\mathbb E(X)=\dfrac{64}{127}f'\left(\dfrac12\right)=\dfrac{64}{127}\times\dfrac{247}{64}=\dfrac{247}{127}$

d.
$\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$

We find $\mathbb E(X^2)$ in a similar manner as in (c).

$\mathbb E(X^2)=\displaystyle\sum_{x=1}^7 x^2\,f_X(x)=\frac{32}{127}\sum_{x=1}^7x^2\left(\frac12\right)^{x-2}$

Now,

$f(y)=\displaystyle\sum_{x=0}^7y^x$
$\implies f'(y)=\displaystyle\sum_{x=1}^7xy^{x-1}$
$\implies f''(y)=\displaystyle\sum_{x=2}^7x(x-1)y^{x-2}$

We know that

$f''(y)=-\dfrac{42y^8-96y^7+56y^6-2}{(1-y)^3}$
$\implies f''\left(\dfrac12\right)=\dfrac{219}{16}$

We also have

$f''(y)=\displaystyle\sum_{x=2}^7x(x-1)y^{x-2}$
$f''(y)=\displaystyle\sum_{x=2}^7x^2y^{x-2}-\sum_{x=2}^7xy^{x-2}$
$f''(y)=\displaystyle\frac1{y^2}\left(\sum_{x=2}^7x^2y^x-\sum_{x=2}^7xy^x\right)$
$f''(y)=\displaystyle\frac1{y^2}\left(\bigg(\sum_{x=1}^7x^2y^x-y\bigg)-\bigg(\sum_{x=1}^7xy^x-y\bigg)\right)$
$f''(y)=\displaystyle\frac1{y^2}\left(\sum_{x=1}^7x^2y^x-\sum_{x=1}^7xy^x\right)$

so that when $y=\dfrac12$ , we get

$\dfrac{219}{16}=4\left(\dfrac{127}{128}\mathbb E(X^2)-\dfrac{127}{128}\mathbb E(X)\right)\implies\mathbb E(X^2)=\dfrac{685}{127}$

Then

$\mathbb V(X)=\dfrac{685}{127}-\left(\dfrac{247}{127}\right)^2=\dfrac{25,986}{16,129}$

6 0

3 years ago