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3 years ago

Find (-6y-2)+5(3+2.5y)

Mathematics

2 answers:

Goshia [24]3 years ago

6 0

I don’t know which one is right but I got 6.5y+13

valentina_108 [34]3 years ago

4 0

The answer is 26-37y over 2. I worked it out and i hope it is very helpful. I know a lot of struggle with math so hope it helps!

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One of the many concerns of college students today is the time required for them to graduate with a baccalaureate degree (BA or

uysha [10]

Answer:

The correct option is C.

Step-by-step explanation:

Given information:

$\overline{x}_1=4.95$

$\overline{x}_2=5.22$

$s_1=0.64$

$s_2=0.89$

$n_1=n_2=21$

The formula for t test statistic t* is

$t^*=\frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$

Using the formula we get

$t^*=\frac{\left(4.95-5.22\right)-(0)}{\sqrt{\frac{0.64^2}{21}+\frac{0.89^2}{21}}}$

$t^*=-1.12869182873$

$t^*\approx -1.13$

The value of test statistic t* is -1.13. Therefore the correct option is C.

6 0

3 years ago

I really do not get this at all and need help please

sweet [91]

I would say C hope you get it correct

4 0

3 years ago

Read 2 more answers

The sample and h is the half-life, in days, of the substance. A sample contains 60% of its original amount of Fermium-257. The h

skelet666 [1.2K]

Answer:

74 days

Step-by-step explanation:

The computation is shown below;

The proportion that remains left after d days is

p = (1/2)^(d ÷ 100)

In the case when the proportion is 60%

So,

0.60 = 0.50^(d ÷ 100)

Now

log(.60) = (d ÷ 100)log(.50)

100·log(.60) ÷ log(.50) = d = 74 days

5 0

3 years ago

PLEASE HELP!!!

kaheart [24]

Hello,

to find other side ;

a = √ hypo^2- leg^2
so, a = √19^2-12^2

therefore it is 14.7

Hope this helps.

7 0

3 years ago

Read 2 more answers

F the voltage V in an electric circuit is held constant, the current I is inversely proportional to the resistance R. If the cu

joja [24]

The current in the 2nd case is 102 A

Step-by-step explanation:

The relationship between voltage, current and resistance in an electric circuit is given by Ohm's law:

$V=IR$

where

V is the voltage

I is the current

R is the resistance

In this problem we are given two sets of data for the current and the resistance, while the voltage is kept constant. This means that we can write:

$V_1=V_2\\I_1 R_1 = I_2 R_2$

where

$I_1 = 60 A$ is the current in the 1st case

$R_1 = 255 \Omega$ is the resistance in the 1st case

$R_2 = 150 \Omega$ is the resistance in the 2nd case

Solving for $I_2$ , we find the current in the 2nd case:

$I_2 = \frac{I_1 R_1}{R_2}=\frac{(60)(255)}{150}=102 A$

Learn more about current and voltage:

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brainly.com/question/10597501

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#LearnwithBrainly

8 0

4 years ago