Round 342 to the nearest ten

The number of purses a vendor sells daily has the probability distribution represented in the table. Number of Purses, x 0 1 2 3

Arada [10]

Answer:

(D)$81

Step-by-step explanation:

Given that the number of purses a vendor sells daily has the probability distribution represented in the table.

Expected Value, $E(x)=\sum_{i=1}^{n}x\cdot p(x)$

Therefore:

$E(x)=(0X0.35)+(1X0.15)+(2X0.2)+(3X0.2)+(4X0.03)+(5X0.07)\\=0+0.15+0.4+0.6+0.12+0.35\\E(x)=1.62$

If each purse sells for $50.00, the number of expected daily total dollar amount taken in by the vendor from the sale of purses

=Expected Value X $50

=1.62 X $50

=$81

The correct option is D.

6 0

3 years ago

Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val

Cloud [144]

Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

$A(t) = Pe^{rt}$

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.

$330 in an account in the year 1999

This means that $P = 330$

$590 in the year 2007

2007 is 8 years after 1999, so P(8) = 590.

We use this to find r.

$A(t) = Pe^{rt}$

$590 = 330e^{8r}$

$e^{8r} = \frac{590}{330}$

$e^{8r} = 1.79$

Applying ln to both sides:

$\ln{e^{8r}} = \ln{1.79}$

$8r = \ln{1.79}$

$r = \frac{\ln{1.79}}{8}$

$r = 0.0726$

Determine the value of the account, to the nearest dollar, in the year 2009.

2009 is 10 years after 1999, so this is A(10).

$A(t) = 330e^{0.0726t}$

$A(10) = 330e^{0.0726*10} = 682$

The value of the account in the year 2009 will be $682.

4 0

3 years ago

How do I work out 1 over 4 + 3 over 8

Dmitrij [34]

You would do 1/4 and get the answer plus 3/8s answer

6 0

3 years ago

In a right triangle one leg measures a quarter of the other, how much do its acute angles measure? with procedure please

KATRIN_1 [288]

Answer:

The answer to your question is 14° and 76°

Step-by-step explanation:

Data

leg 1 = x

leg 2 = x/4

angle 1 = ?

angle 2 = ?

Process

To solve problem use trigonometric functions. We must use the trigonometric function that relates both legs (tangent).

tangent Ф = Opposite side / Adjacent side

-Substitution

tan Ф = (x/4) / x

-Simplification

tan Ф = x / 4x

tan Ф = 1/4

- Find Ф

Ф = tan⁻¹(1/4) = 14°

- Find the other acute angle

The sum of the internal angles in a triangles equals 180°

Ф + Ф₁ + 90° = 180

-Solve for Ф₁

Ф₁ = 180 - 90 - 14

-Result

Ф₁ = 76°

3 0

3 years ago

The water level in a lake decreased by 2 1/4 inches in march and increased by 1 5/8 inches in april. Determine wether the water

mezya [45]

Change in level is given by :

C = increase in level - decrease in level.

Now,

Decrease in level in March is 2 1/4 = 9/4 inches.

Increase in level in April is 1 5/8 = 13/8 inches.

Putting value of these in above equation, we get :

C = 13/8 - 9/4

C = 13/8 - 18/8

C = -5/8 inches

Therefore, the overall level after April will decrease by 5/8 inches.

Hence, option 4. is correct.

3 0

3 years ago