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3 years ago

8

If amy earns 237.60 after working 33 hours what is the hourly rate

1 answer:

lana [24]3 years ago

7 0

To find the hourly rate, divide the total amount earned by total number of hours worked:

237.60 / 33 = 7.2

The hourly rate is $7.20

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In the problem 10 take away 4 equals 6 what is the correct term for the number 4

crimeas [40]

10 would be called the Minuend.
4 would be called the Subtrahend.
4 would be the Difference.

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3 years ago

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Of 10 test scores, eight are less than or equal to 82. What is the percentile rank of a test score of 82?

sdas [7]

Answer:

80%

Step-by-step explanation:

Percentile rank = (number of scores lower or equal to 82 / total number of scores) * 100%

Percentile = (8 / 10) * 100%

= 0.8 * 100%

= 80%

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3 years ago

If y has moment-generating function m(t) = e 6(e t −1) , what is p(|y − µ| ≤ 2σ)?

Lostsunrise [7]

Since $\mathrm M_Y(t)=e^{6(e^t-1}$ , we know that $Y$ follows a Poisson distribution with parameter $\lambda=6$ .

Now assuming $\mu,\sigma$ denote the mean and standard deviation of $Y$ , respectively, then we know right away that $\mu=6$ and $\sigma=\sqrt6$ .

So,

$\mathbb P(|Y-\mu|\le2\sigma)=\mathbb P(6-2\sqrt6\le Y\le6+2\sqrt6)=\dfrac{66366}{175e^6}\approx0.940028$

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4 years ago

When you are trying to find the endpoint when given the midpoint and one point, what would the new formula be? Manipulate the fo

MariettaO [177]

Answer:

$(x_2,y_2) = (2x_m - x_1 ,2y_m - y_1 )$

Step-by-step explanation:

Given

$\left(\frac{x_1+\:x_2}{2},\:\:\frac{y_{1\:}+y_2}{2}\right)\:=\:\left(x_m,\:y_m\right)$

Required

Determine x2, y2

Start by splitting the expression

$x_m = \left(\frac{x_1+\:x_2}{2})$ and $y_m = (\frac{y_{1\:}+y_2}{2})$

Solving for x2 in $x_m = \left(\frac{x_1+\:x_2}{2})$

Multiply through by 2

$2 * x_m = \frac{x_1 + x_2}{2} * 2$

$2x_m = x_1 + x_2$

Make x2 the subject;

$x_2 = 2x_m - x_1$

Similarly:

$y_m = (\frac{y_{1\:}+y_2}{2})$

Multiply through by 2

$2 * y_m = \frac{y_1 + y_2}{2} * 2$

$2y_m = y_1 + y_2$

Make y2 the subject;

$y_2 = 2y_m - y_1$

Hence:

$(x_2,y_2) = (2x_m - x_1 ,2y_m - y_1 )$

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3 years ago

A basin is filled by two pipes in 12 minutes and 16 minutes respectively. Due to the obstruction of water flow after the two pip

olasank [31]

Answer:

The time duration of the two pipes restricted flow before the flow became normal is 4.5 minutes

Step-by-step explanation:

The given information are;

The time duration for the volume, V, of the basin to be filled by one of the pipe, A, = 12 minutes

The time duration for the volume, V, of the basin to be filled by the other pipe, B, = 16 minutes

Therefore, the flow rate of pipe A = V/12

The flow rate of pipe B = V/16

Due to the restriction, we have;

The proportion of its carrying capacity the first pipe, A, carries = 7/8 of the carrying capacity

The proportion of its carrying capacity the second pipe, B, carries = 5/6 of the carrying capacity

Whereby the tank is filled 3 minutes after the restriction is removed, we have;

$\dfrac{7}{8} \times \dfrac{V}{12} \times t + \dfrac{5}{6} \times \dfrac{V}{16} \times t + \dfrac{V}{12} \times 3 + \dfrac{V}{16} \times 3 = V$

Simplifying gives;

$\dfrac{(2\cdot t +7) \cdot V}{16} = V$

2·t + 7 = 16

t = (16 - 7)/2 = 4.5 minutes

Therefore, it took 4.5 seconds of the restricted flow before the the flow of water in the two pipes became normal

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4 years ago

Read 2 more answers