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tamaranim1 [39]
2 years ago
11

7526.442 rounded to the nearest hundredth

Mathematics
1 answer:
AlladinOne [14]2 years ago
3 0
Number > 5 - round up
Number = 5 - round up
Number < 5 - round and keep the number the same

7526.442

Hundredth = 2nd decimal place = 4

2 < 5 
Round and keep the number the same

7526.44
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Geometry: Line QS bisects angle PQR. Solve for x and find the measure of angle PQR.
Kazeer [188]

It is given in the question that,

Line QS bisects angle PQR. Solve for x and find the measure of angle PQR.

And

m \angle PQS = 2x+1&#10;\\&#10;m \angle RQS = 4x-15

Since QS bisects angle PQR, therefore

m \angle PQS = m \angle RQS

Substituting the values, we will get

2x+1=4x-15&#10;\\&#10;1+15 = 4x-2x&#10;\\&#10;2x = 16&#10;\\&#10;x = 8

6 0
1 year ago
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Jake went on a riverboat tour. After returning from the tour, he was curious to know the speed at which the boat was going. Duri
atroni [7]
I'm sorry I'm too lazy to read all of that. However, I did get the main point. The river was going at 3 mph. Since he wanted to travel 10 miles upstream and downstream, you would have to subtract 20 by 3. This would mean he was going at a speed of 17 mph. 
4 0
2 years ago
A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win $1.10; if
topjm [15]

Answer:

a) The expected value is \frac{-1}{15}

b) The variance is  \frac{49}{45}

Step-by-step explanation:

We can assume that both marbles are withdrawn at the same time. We will define the probability as follows

#events of interest/total number of events.

We have 10 marbles in total. The number of different ways in which we can withdrawn 2 marbles out of 10 is \binom{10}{2}.

Consider the case in which we choose two of the same color. That is, out of 5, we pick 2. The different ways of choosing 2 out of 5 is \binom{5}{2}. Since we have 2 colors, we can either choose 2 of them blue or 2 of the red, so the total number of ways of choosing is just the double.

Consider the case in which we choose one of each color. Then, out of 5 we pick 1. So, the total number of ways in which we pick 1 of each color is \binom{5}{1}\cdot \binom{5}{1}. So, we define the following probabilities.

Probability of winning: \frac{2\binom{5}{2}}{\binom{10}{2}}= \frac{4}{9}

Probability of losing \frac{(\binom{5}{1})^2}{\binom{10}{2}}\frac{5}{9}

Let X be the expected value of the amount you can win. Then,

E(X) = 1.10*probability of winning - 1 probability of losing =1.10\cdot  \frac{4}{9}-\frac{5}{9}=\frac{-1}{15}

Consider the expected value of the square of the amount you can win, Then

E(X^2) = (1.10^2)*probability of winning + probability of losing =1.10^2\cdot  \frac{4}{9}+\frac{5}{9}=\frac{82}{75}

We will use the following formula

Var(X) = E(X^2)-E(X)^2

Thus

Var(X) = \frac{82}{75}-(\frac{-1}{15})^2 = \frac{49}{45}

7 0
2 years ago
The amount of bacteria in a Petri dish increases at a rate proportional to the amount present. At time t=0, the amount of bacter
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Answer:

C. 270 grams

Step-by-step explanation:

-This is an exponential growth function which can be expressed using the formula:

P_t=P_oe^{rt}

Where:

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Given the size at t=2 and at t=0, we substitute in the growth function to solve for r:

30=10e^{2r}\\\\3=e^{2r}\\\\r=\frac{In \ 3}{2}\\\\=0.54931

We use this calculated rate to determine the population at t=6

P_t=P_oe^{rt}\\\\=10e^{6\times \frac{In \ 3}{2}}\\\\=270\ grams

Hence, the bacteria's size at t=6 is 270 grams

8 0
1 year ago
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A. No matter what M is, it will correspond to the squared version of 18m
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