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aev [14]
3 years ago
6

Find the value of w and then x. Round lengths to the nearest tenth and angle measures to the nearest degree.

Mathematics
2 answers:
VARVARA [1.3K]3 years ago
4 0

Answer

w=7.6

x=43.7

Step-by-step explanation:

To solve for w you can use Sin(x)=\frac{Opposite}{Hypotenuse}  which would basically be Sin(50)=\frac{w}{10} then solve. Multiply the ten on both sides so you have 10*Sin(50)=w and your final answer is 7.6

To solve for x you can also use Sin(x)=\frac{7.6}{11} then just use Inverse of Sin^-1

The answer is x=43.7

valentina_108 [34]3 years ago
4 0

Answer:

Step-by-step explanation:

The diagram contains 2 right angle triangles.

To determine w, we would apply the the Sine trigonometric ratio. It is expressed as

Sin θ = opposite side/hypotenuse.

Hypotenuse = 10

Opposite side = w

Therefore

Sin 50 = w/10

w = 10Sin50 = 10 × 0.7660

w = 7.66

To determine angle x, we would also apply the sine trigonometric ratio. Therefore

Hypotenuse = 11

Opposite side = 7.66

Sin x = 7.66/11 = 0.6964

x = Sin^-1 (0.6964)

x = 41.1 degrees to the nearest tenth.

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pickupchik [31]

Answer:

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

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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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