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2 years ago

PLEASE HELP!!!

Mathematics

1 answer:

Oksanka [162]2 years ago

4 0

Answer:

PLEASE HELP!!!

1. Under what conditions must we assume a Student t distribution for the sampling distribution of sample means when testing a claim about a population mean?

2. Give one difference between the Student t distribution and the normal distribution.

3. Which TI-84 calculator command or StatCrunch dialog box is used to find the P-value given a t test statistic?

Step-by-step explanation:

PLEASE HELP!!!

1. Under what conditions must we assume a Student t distribution for the sampling distribution of sample means when testing a claim about a population mean?

2. Give one difference between the Student t distribution and the normal distribution.

3. Which TI-84 calculator command or StatCrunch dialog box is used to find the P-value given a t test statistic?

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2c +3 = 3c - 4 Help me please

VashaNatasha [74]

Hey there!

Solve the equation:

Answer :

c = 7 ✅

Explanation

2c + 3 = 3c - 4

>> Subtract 3 from both sides :

2c + 3 - 3 = 3c - 4 - 3

2c = 3c - 7

>> Substrat 3c from both sides :

2c - 3c = 3c - 7 - 3c

-c = -7

>> Divide each side by -1 :

-c / -1 = -7 / -1

c = 7

Let's verify:

2c + 3 ⇔ 2(7) + 3 ⇔ 14 + 3 ⇔ 17

3c - 4 ⇔ 3(7) - 4 ⇔ 21 - 4 ⇔ 17

Therefore, your answer is c = 7 .

Mor about equation :

brainly.com/question/27353929

Have a good day :)

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2 years ago

Read 2 more answers

Help help help help please guys

maria [59]

I think the answer is 9
It's a smaller version of the other figure.
21/3=7
27/3=9

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3 years ago

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Rotate 90 degrees clockwise around the origin and then translate down. Reflect across the x-axis and then reflect across the y-axis.

Step-by-step explanation:

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2 years ago

If 13 = 4.5 -5k what is the value of k

USPshnik [31]

Answer:

K = negative 1.7

Step-by-step explanation:

4.5- 13 = -8.5

-8.5/5 = -1.7

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Hope this helped

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3 years ago

Evaluate the sum of the following finite geometric series.

rjkz [21]

Answer:

$\large\boxed{\dfrac{156}{125}\approx1.2}$

Step-by-step explanation:

<h3>Method 1:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\\\\for\ n=1\\\\\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1\\\\for\ n=2\\\\\left(\dfrac{1}{5}\right)^{2-1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}\\\\for\ n=3\\\\\left(\dfrac{1}{5}\right)^{3-1}=\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\\\for\ n=4\\\\\left(\dfrac{1}{5}\right)^{4-1}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}$

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}=\dfrac{125}{125}+\dfrac{25}{125}+\dfrac{5}{125}+\dfrac{1}{125}=\dfrac{156}{125}$

<h3>Method 2:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\to a_n=\left(\dfrac{1}{5}\right)^{n-1}\\\\\text{The formula of a sum of terms of a geometric series:}\\\\S_n=a_1\cdot\dfrac{1-r^n}{1-r}\\\\r-\text{common ratio}\to r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=\left(\dfrac{1}{5}\right)^{n+1-1}=\left(\dfrac{1}{5}\right)^n\\\\r=\dfrac{\left(\frac{1}{5}\right)^n}{\left(\frac{1}{5}\right)^{n-1}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\r=\left(\dfrac{1}{5}\right)^{n-(n-1)}=\left(\dfrac{1}{5}\right)^{n-n+1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}$

$a_1=\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1$

$\text{Substitute}\ a_1=1,\ n=4,\ r=\dfrac{1}{5}:\\\\S_4=1\cdot\dfrac{1-\left(\frac{1}{5}\right)^4}{1-\frac{1}{5}}=\dfrac{1-\frac{1}{625}}{\frac{4}{5}}=\dfrac{624}{625}\cdot\dfrac{5}{4}=\dfrac{156}{125}$

5 0

3 years ago