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Anon25 [30]
4 years ago
6

Solve for x using figure to the right. Pls answer this it is urgent

Mathematics
1 answer:
goldenfox [79]4 years ago
8 0

Answer:

Therefore the value of x = 10 units

Step-by-step explanation:

Let label the Triangles first,

Δ ABC a right triangle at ∠ A =90°

Δ ADB andΔ ADC a right triangle at ∠ D =90°

Such that

AD = x

BD = 50

CD = 2

∴ BC = BD + DC = 50 + 2 = 52

To Find:

x = ?

Solution:

In right triangle  By Pythagoras Theorem,

(\textrm{Hypotenuse})^{2} = (\textrm{Shorter leg})^{2}+(\textrm{Longer leg})^{2}

In right triangle Δ ADB andΔ ADC  By Pythagoras Theorem we will have,

AB² =  BD² + AD²

AB² = 50² + x²            ..................equation ( 1 )

and

AC² = DC² + AD²

AC² = 2² + x²             ...................equation ( 2 )

Now in right triangle Δ ABC,

BC² = AB² + AC²

Equating equation  (1 ) and ( 2 ) and the given value we get

52² = 50² + x² + 2² + x²

∴ 2x² = 2704 - 2504

∴ 2x² =200

∴ x^{2} =\frac{200}{2}\\\\\therefore x=\pm\sqrt{100} \\\\ \textrm{x cannot be negative}\\\therefore x= 10\ unit

Therefore the value of x = 10 units

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A line passes through (10,2) and (14,-22). Write the equation of the line in standard form.
jarptica [38.1K]

Answer:

The equation of the line in standard form is:

6x + y = 62

Step-by-step explanation:

Given the points

  • (10,2)
  • (14,-22)

Determining the slope between (10, 2) and (14, -22)

\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}

\left(x_1,\:y_1\right)=\left(10,\:2\right),\:\left(x_2,\:y_2\right)=\left(14,\:-22\right)

m=\frac{-22-2}{14-10}

m=-6

The point-slope form of the line equation is  

y-y_1=m\left(x-x_1\right)

where

  • m is the slope of the line
  • (x₁, y₁) is the point

substituting the values m = -6 and the point (10, 2) in the point-slope form of the line equation

y-y_1=m\left(x-x_1\right)

y - 2 = -6(x - 10)

y - 2 = -6x +60

adding 2 to both sides

y-2+2 = -6x + 60 + 2

y = -6x +  62

We can write the equation in the standard form such as

Ax + By = C

Thus,

y = -6x +  62

adding -6x to both sides

6x + y = -6x + 62 + -6x

6x + y = 62

Therefore, the equation of the line in standard form is:

6x + y = 62

4 0
3 years ago
Listed below are the number of years it took for a random sample of college students to earn bachelor's degrees (based on data f
schepotkina [342]

Answer:

a) Mean = 6.5, sample standard deviation = 3.50

b) Standard error = 0.7826

c) Point estimate = 6.5

d) Confidence interval:  (5.1469 ,7.8531)

Step-by-step explanation:

We are given the following data set for students to earn bachelor's degrees.

4, 4, 4, 4, 4, 4, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 6, 6, 8, 9, 9, 13, 13, 15

a) Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{130}{20} = 6.5

Sum of squares of differences = 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 4 + 4 + 4 + 4 + 4 + 4 + 0.25 + 0.25 + 2.25 + 6.25 + 6.25 + 42.25 + 42.25 + 72.25 = 233.5

S.D = \sqrt{\frac{233.5}{19}} = 3.50

b) Standard Error

= \displaystyle\frac{s}{\sqrt{n}} = \frac{3.50}{\sqrt{20}} = 0.7826

c) Point estimate for the mean time required for all college is given by the sample mean.

\bar{x} = 6.5

d) 90% Confidence interval:  

\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}  

Putting the values, we get,  

t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.10} = \pm 1.729  

6.5 \pm 1.729(\frac{3.50}{\sqrt{20}} ) = 6.5 \pm 1.3531 = (5.1469 ,7.8531)

e) No, the confidence interval does not contain the value of 4 years. Thus, confidence interval is not a good estimator as most of the value in the sample is of 4 years. Most of the sample does not lie in the given confidence interval.

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