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IRISSAK [1]
3 years ago
11

2x + 5y = 19 y = 3x - 3 Solve for x and y

Mathematics
1 answer:
ozzi3 years ago
8 0

I dont know if i did this right but here

x1 = 9.5    y1 = 0

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What equivalent to 5m(m+3)
Lostsunrise [7]

Answer:

5m^2 + 15m

Step-by-step explanation:

use distributive property:

5m(m) is 5m^2

5m(3) is 15m

which is 5m^2 + 15m

3 0
3 years ago
To find probability, divide the number of __________ by the number of trials
seropon [69]

Answer:

To find probability, divide the number of times an event happens by the number of trials.

4 0
3 years ago
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Test the claim that the mean GPA of night students is larger than 3.1 at the .10 significance level. The null and alternative hy
Alborosie

Answer:

H 0 : μ = 3.1    H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66

df = n-1=75-1 =74

The p value is:

p_v = P(t_{74} >8.66) =3.65x10^{-13}

The significance level is: 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

Reject the null hypothesis

Step-by-step explanation:

For this case we want to test the claim that mean GPA of night students is larger than 3.1 at the .10 significance level. The claim needs to be on the alternative hypothesis so then we have the following system of hypothesis:

H 0 : μ = 3.1    H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

We have the following info given:

\bar X = 3.13 , s =0.03 , n =75

The statistic to check the hypothesis is given by:

t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}

Replacing the info given we got:

t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66

The degrees of freedom are given by:

df = n-1=75-1 =74

The p value since is a right tailed ted is given by:

p_v = P(t_{74} >8.66) =3.65x10^{-13}

The significance level is 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

Reject the null hypothesis

6 0
3 years ago
A 30-N stone is dropped from a height of 10 m and strikes the ground with a speed of 13 m/s.
Anon25 [30]

Answer:

4.1 N

Step-by-step explanation:

We can solve this problem by using considerations about energy.

At the moment the stone is dropped, it has only gravitational potential energy:

PE=mgh

where

mg=30 N is the weight of the stone

h = 10 m is the initial height of the stone

As the stone falls, part of this energy is converted into kinetic energy, while part into thermal energy due to the presence of the air friction, acting opposite to the motion of the stone:

KE+E_t = \frac{1}{2}mv^2 + E_t

where:

m=\frac{mg}{g}=\frac{30}{9.8}=3.06 kg is the mass

v = 13 m/s is the final speed of the stone

E_t is the thermal energy

The thermal energy is actually equal to the work done by the air friction on the stone:

E_t=W=Fh

where

F is the average force of friction

h = 10 m

Since the total energy must be conserved, we can combine the three equations, so we find:

mgh=\frac{1}{2}mv^2+Fh

And solving for F, we find the average force of air friction:

F=\frac{mgh-0.5mv^2}{h}=\frac{(30)(10)-0.5(3.06)(13)^2}{10}=4.1 N

3 0
3 years ago
PLEASE HELP!!! ILL GIVE BRAINLIEST
MakcuM [25]

Answer:

*gives help*

:)

Step-by-step explanation:

7 0
3 years ago
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