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3 years ago

Math can you please help me with this thank you!

Mathematics

1 answer:

padilas [110]3 years ago

4 0

Tip: whenever y equals a number, the line is horizontal and the slope is 0. if the number equals x, the line will be vertical and there is no slope.

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Round 30713.517126 to the nearest hundred.

larisa86 [58]

Answer:

3,266.53

Step-by-step explanation:

7 0

3 years ago

The average density of the material in intergalactic space is approximately 2.5 × 10–27 kg/m3. what is the volume of a lead samp

Lesechka [4]

Answer:

e. $1.8\times 10^{-6}m^3$

Step-by-step explanation:

It is given that,

The density of intergalactic space material is $2.5 \times 10^{-27}$ kg per cubic meter.

And the volume of intergalactic space material is $8.0\times 10^{24} m^3$

So the mass of that much intergalactic space material is,

$m= \rho\times v$ , where 'm' is the mass, and 'v' is the volume.

Putting the values we get,

$m=2.5 \times 10^{-27}\times 8.0 \times 10^{24}$

$m=20\times 10^{(-27+24)}=20\times 10^{-3} kg$

It is also given that the mass of lead is the same as the mass of the intergalactic space material. Therefore, the mass of lead is $20 \times 10^{-3}=2 \times 10^{-2}kg$

So the volume of lead is,

$v=\frac{m}{\rho} =\frac{2 \times 10^{-2}}{11300} = 0.00000177 m^3$

$v=1.77 \times 10^{-6}=1.8\times 10^{-6}m^3$

So the volume of lead is $v=1.8\times 10^{-6}m^3$ .

3 0

3 years ago

V1/V2=T1/T2. Solve for V1

Ahat [919]

V1=(T1/T2)*V2
In this problem, since you are given no actual numbers, the only thing you can do is multiply V2 over to get V1 alone. Since you can't cancel anything out, you are left with the answer above.

4 0

3 years ago

Find the slope of the following equation .simplify your answer 5x+2y=-10

defon

Answer:

m=-5/2

Step-by-step explanation:

Use the slope-intercept form y=mx+b to find the slope m.

6 0

3 years ago

Read 2 more answers

The score on an exam from a certain MAT 112 class, X, is normally distributed with μ=78.1 and σ=10.8.

salantis [7]

a) $X$

b) 0.1539

c) 0.1539

d) 0.6922

Step-by-step explanation:

In this problem, the score on the exam is normally distributed with the following parameters:

$\mu=78.1$ (mean)

$\sigma = 10.8$ (standard deviation)

We call X the name of the variable (the score obtained in the exam).

Therefore, the event "a student obtains a score less than 67.1) means that the variable X has a value less than 67.1. Mathematically, this means that we are asking for:

$X$

And the probability for this to occur can be written as:

$p(X$

To find the probability of X to be less than 67.1, we have to calculate the area under the standardized normal distribution (so, with mean 0 and standard deviation 1) between $z=-\infty$ and $z=Z$ , where Z is the z-score corresponding to X = 67.1 on the s tandardized normal distribution.

The z-score corresponding to 67.1 is:

$Z=\frac{67.1-\mu}{\sigma}=\frac{67.1-78.1}{10.8}=-1.02$

Therefore, the probability that X < 67.1 is equal to the probability that z < -1.02 on the standardized normal distribution:

$p(X$

And by looking at the z-score tables, we find that this probability is:

$p(z$

And so,

$p(X$

Here we want to find the probability that a randomly chosen score is greater than 89.1, so

$p(X>89.1)$

First of all, we have to calculate the z-score corresponding to this value of X, which is:

$Z=\frac{89.1-\mu}{\sigma}=\frac{89.1-78.1}{10.8}=1.02$

Then we notice that the z-score tables give only the area on the left of the values on the left of the mean (0), so we have to use the following symmetry property:

$p(z>1.02) =p(z$