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Yanka [14]
3 years ago
15

Identify the slope and y-intercept of the graph of the equation.

Mathematics
1 answer:
IceJOKER [234]3 years ago
5 0

Answer:

c.

Step-by-step explanation:

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kakasveta [241]
(-1.5,2.5) is the point where A is on the coordinate plane.
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3 years ago
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6x-9-2(1-x)=X+9 solve this​
Kay [80]
  • Answer:

<em>x = 20/7</em>

<em>x = 2  6/7</em>

  • Step-by-step explanation:

<em>Hi there !</em>

6x - 9 - 2(1 - x) = x + 9

<em>6x - 9 - 2 + 2x = x + 9</em>

<em>6x + 2x - x = 9 + 9 + 2</em>

<em>8x - x = 20</em>

<em>7x = 20</em>

<em>x = 20/7</em>

<em>x = 2  6/7</em>

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<em>Good luck !</em>

3 0
3 years ago
Question 11 of 12
iogann1982 [59]

Answer:

By ASA

Step-by-step explanation:

<SVR=<VRT (ALTERNATE ANGLES)

SIDE RV=SIDE RV (COMMON)

<SRV=<TVR (ALTERNATE ANGLES)

So by side angle side postulate both triangles are congruent

Hope u understand

Brainliest?

6 0
3 years ago
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Nissan wants to administer a satisfaction survey to its current customers. Using their customer​ database, the company randoml
Irina-Kira [14]

Answer: simple random sampling

Step-by-step explanation:

  • A simple random sampling is a sampling method in which individuals are chosen randomly from the population. In this each individual has equal chances to get selected.
  • If N is the population size , then for any individual , the chances for getting selected is \dfrac{1}{N}.

As per given , the company randomly selects 50 customers directly from the database without applying any further steps and asks them about their level of satisfaction with the company.

Thus , the sampling technique used here is simple random sampling.

3 0
4 years ago
18. A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be
avanturin [10]

Answer:

a. =50 ± 4.67

b. =50 ± 4.81  

decreasing the sample size increase the margin of error

c. =50 ± 3.51

decreasing the confidence level reduced the margin of error

d. No.  because the confidence interval coefficient is gotten from the table of normal distribution of Z value.  therefore the confidence interval is only used for normal distributed population

Step-by-step explanation:

given

The sample mean, x, = 50,

the sample standard deviation, s, = 8.

a. Construct a 98% confidence interval for m if the sample size, n, is 20

for a 98% confidence interval of an infinite population, thus for a sample of size N and mean x, drawn from an infinite population having a standard deviation of σ, the mean value of the population is estimated to by:

x± 2.33σ /√N

for a confidence level of 98%, Zc = 2.33 gotten from the table of confidence interval.

This indicates that the mean value of the population lies between

50- ( 2.33*8 /√20)      and 50+ ( 2.33*8 /√20)

with 98% confidence in this prediction.

=50 ± 4.67

=45.33 or 54.67

(b) Construct a 98% confidence interval for m if the sample size, n, is 15.

using the same method as above

x± 2.33σ /√N

This indicates that the mean value of the population lies between

50- ( 2.33*8 /√15)      and 50+ ( 2.33*8 /√15)

=50 ± 4.81

=45.19 or 54.81

decreasing the sample size increase the margin of error

(c) Construct a 95% confidence interval for m if the sample size, n, is 20. Compare the results to those obtained in part

95% confidence interval =1.96 (gotten from table of confidence coefficient)

using x±1.96 σ /√N

This indicates that the mean value of the population lies between

50- (1.96 *8 /√20)      and 50+ (1.96 *8 /√20)

=50 ± 3.51

=46.49 or 53.51

c. decreasing the confidence level reduced the margin of error

(d) Could we have computed the confidence intervals in parts (a)–(c) if the population had not been normally distributed?

No.  because the confidence interval coefficient is gotten from the table of normal distribution of Z value.  therefore the confidence interval is only used for normal distributed population

7 0
3 years ago
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