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3 years ago

Can someone help me and show your work thanks

Mathematics

2 answers:

GREYUIT [131]3 years ago

4 0

Answer:

D) $8.20

Step-by-step explanation:

82 x 0.1 = 8.2

gregori [183]3 years ago

3 0

Answer: $8.20

Step-by-step explanation:

Take 82 and multiply it by 0.10

You can use this equation

82(n)

n will equal 0.10

82(0.10)

$8.20

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Answer the problem and show all work.

marishachu [46]

Member Price = 30 + 3x

Non-member Price = 6x

x = the tickets they buy.

We want 30 + 3x = 6x

So, first we subtract 3x from each side and are left with:

30 = 3x

Then, divide each side by 3.

10 = x

So, the cost of 10 tickets is the same more non-members and members.

We can also check it:

Member Price: 30 + 3(10)

30 + 30 = 60

Non-member Price: 6(10)

8 0

2 years ago

The graph of the linear function is shown on the coordinate grid. What is the equation of this line?

Serjik [45]

Y=-1/3-2

Hope this helps!! :)

6 0

2 years ago

Evaluate the expression t-6/4 for t =38

Finger [1]

Answer:

Exact Form:

73 over 2

Decimal Form:

36.5

Mixed Number Form:

36 1 over 2

Step-by-step explanation:

Substitute the value of the variable into the equation and simplify.

8 0

3 years ago

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

2 years ago

Assume that we have two events, and , that are mutually exclusive. Assume further that we know and . If an amount is zero, enter

sweet [91]

Answer:

Explained below.

Step-by-step explanation:

The complete question is:

Assume that we have two events, A and B, that are mutually exclusive. Assume further that we know P(A) = 0.30 and P(B) =0.40.

What is P(A and B)?

What is P(A | B)?

Is P(A | B) equal to P(A)?

Are events A and B dependent or independent?

A student in statistics argues that the concepts of mutually exclusive events and independent events are really the same, and that if events are mutually exclusive they must be independent. Is this statement accurate?

What general conclusion would you make about mutually exclusive and independent events given the results of this problem?

Solution:

The probability of the two events A and B are:

P(A) = 0.30 and P(B) =0.40

(a)

Compute the value of P (A ∩ B) as follows:

$P(A\cap B)=0$

This is because mutually exclusive events are those events that cannot occur together.

(b)

Compute the value of P (A | B) as follows:

$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{0}{0.40}=0$

Thus, the value of P (A | B) is 0.

(c)

No, P (A | B) is not equal to the P (A).

(d)

As mentioned in part (a), mutually exclusive events are those events which cannot occur together.

That is, $P(A\cap B)=0$ .

Events A and B are independent if the chance of their concurrent happening is equivalent to the multiplication of their distinct probabilities.

That is, $P(A\cap B)=P(A)\times P(B)$ .

The concepts of mutually exclusive events and independent events are not the same.

(e)

As the it is provided that A and B are mutually exclusive events, we know that $P(A\cap B)=0$ .

Now compute the value of $P(A)\times P(B)$ as follows:

$P(A)\times P(B)=0.30\times 0.40=0.12\neq 0$

Thus, the events A and B are not independent.

Thus, if two events are mutually exclusive events they cannot be independent.

4 0

3 years ago