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den301095 [7]
2 years ago
14

The following random sample from a population whose values were normally distributed was collected.10 8 11 11The 95% confidence

interval for μ is:a. 8.00 to 10.00b. 7.75 to 12.25c. 9.75 to 10.75d. 8.52 to 10.98
Mathematics
1 answer:
Margaret [11]2 years ago
6 0

Answer:

Answer:

B

Step-by-step explanation:

First to find the sample mean

Xbar= summation x/n= 40/4= 10

The sample standard deviation is:

S= √summation(x-xbar)²/ n-1

=√6/3= 1.4142

The degrees of freedom

df= N-1= 4-1= 3

For 95% confidence level, critical value of t

t= 3.182

The 95% confidence interval is:

=Xbar plus or minus t' *8/√n

= 10 plus or minus 3.182(1.4142/√4)

= 10plus or minus 2.25

So 7.75 or 12.25

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197 people ride the bus in the morning. 10 less than 84 people ride the bus in the evening. How many people ride the bus in all?
SVETLANKA909090 [29]

Answer:

Step-by-step explanation:

128

7 0
2 years ago
Which quadratic equation is equivalent to (x – 4)2 – (x – 4) – 6 = 0?
Viktor [21]

I hope choices must be given in the problem.

I am showing the method to find the equivalent equation of the above equation. You can match with your given choices.

First step is to expand the first term. So,

(x-4)² = (x - 4)(x - 4) Since a²= a*a

= x² - 4x - 4x + 4 * 4 By multiplying.

= x² -8x + 16 Combine the like terms.

So, (x - 4)² - (x -4) - 6

= x² - 8x + 16 - x + 4 - 6

= x² - 9x + 14 Combine the like terms.

So, equivalent equation of the above equation is x² - 9x + 14 = 0.

5 0
2 years ago
Read 2 more answers
Square $ABCD$ has area $200$. Point $E$ lies on side $\overline{BC}$. Points $F$ and $G$ are the midpoints of $\overline{AE}$ an
Charra [1.4K]

1. Consider square ABCD. You know that  

A_{ABCD}=AD^2=200,

then

AB=BC=CD=AD=10\sqrt{2}.

2. Consider traiangle AED. F is mipoint of AE and G is midpoint of DE, then FG is midline of triangle AED. This means that

FG=\dfrac{AD}{2}=\dfrac{10\sqrt{2} }{2}=5\sqrt{2}.

3. Consider trapezoid BFGC. Its area is

A_{BFGC}=\dfrac{FG+BC}{2}\cdot h, where h is the height of trapezoid and is equal to half of AB. Thus,

A_{BFGC}=\dfrac{FG+BC}{2}\cdot \dfrac{AB}{2}=\dfrac{5\sqrt{2}+10\sqrt{2}}{2}\cdot \dfrac{10\sqrt{2}}{2}=75.

4.

A_{BFGC}=A_{BFGE}+A_{EGC},\\A_{EGC}=A_{BFGC}-A_{BFGE}=75-34=41.

5. Note that angles EGC and CGD are supplementary and

\sin \angle CGD=\sin \angle EGC.

Then

A_{CGD}=\dfrac{1}{2}CG\cdot CD\cdot \sin \angle CGD=\dfrac{1}{2}CG\cdot EG\cdot \sin \angle CGE=A_{ACG}=41.

Answer: A_{CGD}=41.

8 0
2 years ago
Kiran has 27 nickels and quarters in his pocket, worth a total of $2.75.
Montano1993 [528]

The system of equations is

n + d = 27

n + 5d = 55

Step-by-step explanation:

The given is:

  • Kiran has 27 nickels and quarters in his pocket
  • They worth a total of $2.75
  • We need to write a system of equations to represent the relationships between the number  of nickels n, the number of quarters d, and the dollar amount in this situation

∵ The number of nickles is n

∵ The number of quarters is d

∵ There are 27 nickles and quarters

∴ n + d = 27 ⇒ (1)

∵ 1 nickel = 5 cents

∵ 1 quarter = 25 cents

- Multiply n by 5 and d by 25 to find their value

∴ They worth = 5n + 25d

∵ They worth a total of $2.75

- Chang the dollar to cent

∵ 1 dollar = 100 cents

∴ $2.75 = 2.75 × 100 = 275 cents

- Equate their value by 275

∴ 5n + 25d = 275

- Simplify it by divide each term by 5

∴ n + 5d = 55 ⇒ (2)

The system of equations is

n + d = 27

n + 5d = 55

Learn more:

You can learn more about the system of equations in brainly.com/question/2115716

#LearnwithBrainly

5 0
2 years ago
if the Browns plant 3/16 tomatoes, 1/4cabbage, and 2/8pepper, what fraction of their garden will be unplanted?
zysi [14]
5/16 will be unplanted
3/16+4/16+4/16=11/16
16/16-11/16=5/16
6 0
2 years ago
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