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Alexxx [7]
2 years ago
5

If the given ordered pair equal then find value of x and y(3x+5)=(x+2)and (4y+7)= (y+3)​

Mathematics
1 answer:
Yanka [14]2 years ago
7 0

Answer:

3x+5 =x+2

3x-x=2-5

2x= –3

x=-3/2

4y+7=y+3

4y–y =3-7

3y=–4

y= —4/3

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In a survey, 11 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped
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The population standard deviation is not known.

90% Confidence interval by T₁₀-distribution: (38.3, 53.7).

Step-by-step explanation:

The "standard deviation" of $14 comes from a survey. In other words, the true population standard deviation is not known, and the $14 here is an estimate. Thus, find the confidence interval with the Student t-distribution. The sample size is 11. The degree of freedom is thus 11 - 1 = 10.

Start by finding 1/2 the width of this confidence interval. The confidence level of this interval is 90%. In other words, the area under the bell curve within this interval is 0.90. However, this curve is symmetric. As a result,

  • The area to the left of the lower end of the interval shall be 1/2 \cdot (1 - 0.90)= 0.05.
  • The area to the left of the upper end of the interval shall be 0.05 + 0.90 = 0.95.

Look up the t-score of the upper end on an inverse t-table. Focus on the entry with

  • a degree of freedom of 10, and
  • a cumulative probability of 0.95.

t \approx 1.812.

This value can also be found with technology.

The formula for 1/2 the width of a confidence interval where standard deviation is unknown (only an estimate) is:

\displaystyle t \cdot \frac{s_{n-1}}{\sqrt{n}},

where

  • t is the t-score at the upper end of the interval,
  • s_{n-1} is the unbiased estimate for the standard deviation, and
  • n is the sample size.

For this confidence interval:

  • t \approx 1.812,
  • s_{n-1} = 14, and
  • n = 11.

Hence the width of the 90% confidence interval is

\displaystyle 1.812 \times \frac{14}{\sqrt{10}} \approx 7.65.

The confidence interval is centered at the unbiased estimate of the population mean. The 90% confidence interval will be approximately:

(38.3, 53.7).

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