1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Step2247 [10]
3 years ago
15

The weights of four randomly and independently selected bags of potatoes labeled 20.0 pounds were found to be 20.8​, 21.3​, 20.7

​, and 21.2 pounds. Assume Normality. Answer parts​ (a) and​ (b) below. a. Find a​ 95% confidence interval for the mean weight of all bags of potatoes.
Mathematics
1 answer:
Amanda [17]3 years ago
4 0

Answer:

(20.532 , 21.468)

Step-by-step explanation:

Enter data into L₁

CL= 0.95

df= 3

Distribution:  xbar ~ t₃

Calculator commands (for TI only): [Stat],[TESTS],[TInterval],[Data],(List: L₁),(Freq:1),(C-Level:0.95),[Calculate]

Confidence interval (20.532,21.468)

We can conclude that the true population mean for bags of potatoes labeled 20.0 pounds is between (20.532,21,468) with a 95% confidence level.

You might be interested in
CAN smb HELP me please 20 POINTS!!!! and im giving brainly
Savatey [412]

Answer:

a and c

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
What is another way to write the number 271
Murrr4er [49]
Two hundred seventy one 
8 0
3 years ago
Read 2 more answers
2. Explain, in words, how you would solve this problem. Hint: you should get 1 point
makkiz [27]

Answer:

9-24=4(3)

-15=4×3

-15=12

7 0
3 years ago
Read 2 more answers
What is the coefficient of x2y3 in the expansion of (2x + y)5?
Zigmanuir [339]

Option C:

The coefficient of x^{2} y^{3} is 40.

Solution:

Given expression:

(2 x+y)^{5}

Using binomial theorem:

(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}

Here a=2 x, b=y

Substitute in the binomial formula, we get

(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}

                                                            $+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}

Let us solve the term one by one.

$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}

$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y

$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}

$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}

$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}

$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}

Substitute these into the above expansion.

(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}

The coefficient of x^{2} y^{3} is 40.

Option C is the correct answer.

5 0
2 years ago
The members of the boosters organization at your high school bought new balls for the school . They spent $ 22.00 per basketball
azamat
The askser is 283828288288 god it’s da key bro
6 0
3 years ago
Other questions:
  • Write the fraction or decimal as a percent #1. 0.622 ,#2. 0.303 #3. 2.45
    5·1 answer
  • Order the integers {–2, 4, –1, 2, –8} from least to greatest.
    12·2 answers
  • X + (-y)= -y + x<br> What’s the property
    13·2 answers
  • J bought three 15pund bags of dry dog food. Her dog eats 10 ounces of dry dog food a day. After how many days will J need to buy
    10·1 answer
  • 4/5 is equivalent to 12 over what
    11·2 answers
  • 50 POINTS PLEASE HELP ME!! PLEASE HURRY ASAP!!!!!<br> Solve for x
    13·1 answer
  • What value of g makes this equation true: g+4=-17 ?
    13·1 answer
  • I need help pleassse
    6·1 answer
  • K(x)= -3x - 3; find k(-2)
    12·1 answer
  • Evaluate b-(-1/8)+c where b=2 and c=-7/4
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!