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3 years ago

Helppppppppppppppppp

Mathematics

1 answer:

Gwar [14]3 years ago

5 0

Answer:

Step-by-step explanation:

40-10=30 so N has to be 30

Your answers are N=30 and N-25=5

Hope this helps :)

1 thanks for me=1 thanks for you

5 stars=3 thanks for you

Brainliest for me=5 thanks for you

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Suppose PV =NRT, where P=8, V=2X, N=2 and R=2X.you are given X=3. Solve for T

FromTheMoon [43]

Answer: T = 4

Step-by-step explanation:

1. Write all the variables down

P = 8 V = 2X N = 2 R = 2X X = 3

2. Since you know that X = 3 substitute it in to find V and R

V = 2X = 2(3) = 6

R = 2X = 2(3) = 6

3. Find PV

PV = P x V

= 8 x 6

= 48

4. Find NRT

NRT = N x R x T

= 2 x 6 x T

= 12 x T

= 12T

5. Find T

PV = NRT

48 = 12T

12T = 48

divide both sides by 12

T = 48 ÷ 12

T = 4

7 0

2 years ago

Please help with my algebra 1 homework <br><br> thank you

MrRissso [65]

1) Answer:

$\frac{4m^{10} }{9n^{18} }$

2) Answer:

$8m ^{5}$

The answer for first equation is, option B.

and the Answer for Second equation is option D.

6 0

2 years ago

The mean SAT score in mathematics, M, is 600. The standard deviation of these scores is 48. A special preparation course claims

kupik [55]

Answer:

Step-by-step explanation:

The mean SAT score is $\mu=600$ , we are going to call it \mu since it's the "true" mean

The standard deviation (we are going to call it $\sigma$ ) is

$\sigma=48$

Next they draw a random sample of n=70 students, and they got a mean score (denoted by $\bar x$ ) of $\bar x=613$

The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.

- So the Null Hypothesis $H_0:\bar x \geq \mu$

- The alternative would be then the opposite $H_0:\bar x < \mu$

The test statistic for this type of test takes the form

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}$

and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.

With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}\\\\= \frac{| 600-613 |}{48/\sqrt(70}}\\\\= \frac{| 13 |}{48/8.367}\\\\= \frac{| 13 |}{5.737}\\\\=2.266\\$

<h3>since 2.266>1.645 we can reject the null hypothesis.</h3>