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3 years ago

The sides of a rhombus with angle of 60° are 6 inches. find the area of the rhombus. 9√3 in2 18√3 in2 36 in2

Mathematics

2 answers:

musickatia [10]3 years ago

8 0

Drop a perpendicular from an obtuse vertex.

That makes a 30 - 60º right triangle, with hypotenuse 6 and short leg 3. So the long leg, which is the height of the rhombus, is 3√3 .

Since a rhombus is a parallelogram, its area is base x height = 6 x 3√3 = 18√3

harina [27]3 years ago

4 0

Drop a perpendicular from an obtuse vertex.

That makes a 30 - 60º right triangle, with

hypotenuse 6 and short leg 3. So the long leg,

which is the height of the rhombus, is 3√3 .

Since a rhombus is a parallelogram, its area

is base x height = 6 x 3√3 = 18√3 .

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I need help with number 2!! please help!

Maurinko [17]

Answer:

12.68

Step-by-step explanation:

76 degrees /2 = 38

Half the width of 15m forms the lower side of the triangle: 7.5 m

sin(38) = 7.5/h

=12.1820

12.18+.5

12.68

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3 years ago

solve the word problem. paula has 24 coins in her drawer. seven of the coins are nickels and the rest are pennies. what is the t

Arlecino [84]

52 cents
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3 years ago

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x overbar, is found

joja [24]

Answer:

(a) 80% confidence interval for the population mean is [109.24 , 116.76].

(b) 80% confidence interval for the population mean is [109.86 , 116.14].

(c) 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed.

Step-by-step explanation:

We are given that a simple random sample of size n is drawn from a population that is normally distributed.

The sample mean is found to be 113 and the sample standard deviation is found to be 10.

(a) The sample size given is n = 13.

Firstly, the pivotal quantity for 80% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = 113

s = sample standard deviation = 10

n = sample size = 13

$\mu$ = population mean

Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 80% confidence interval for the population mean, $\mu$ is ;

P(-1.356 < $t_1_2$ < 1.356) = 0.80 {As the critical value of t at 12 degree

of freedom are -1.356 & 1.356 with P = 10%}

P(-1.356 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.356) = 0.80

P( $-1.356 \times }{\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.356 \times }{\frac{10}{\sqrt{13} } }$ , $113+1.356 \times }{\frac{10}{\sqrt{13} } }$ ]

= [109.24 , 116.76]

Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].

(b) Now, the sample size has been changed to 18, i.e; n = 18.

So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.

80% confidence interval for $\mu$ = [ $\bar X-1.333 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.333 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.333 \times }{\frac{10}{\sqrt{18} } }$ , $113+1.333 \times }{\frac{10}{\sqrt{18} } }$ ]

= [109.86 , 116.14]

Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].

(c) Now, we have to construct 98% confidence interval with sample size, n = 13.

So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.

98% confidence interval for $\mu$ = [ $\bar X-2.681 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+2.681 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-2.681 \times }{\frac{10}{\sqrt{13} } }$ , $113+2.681 \times }{\frac{10}{\sqrt{13} } }$ ]

= [105.56 , 120.44]

Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed because t test statistics is used only when the data follows normal distribution.

6 0

3 years ago

Which property of polynomial addition says that the sum of two polynomials is always a polynomial

SpyIntel [72]

Maybe it is closure property

5 0

3 years ago

The speed at which cars travel on the highway has a normal distribution with a mean of 60 km/h and a standard deviation of 5 km/

oee [108]

The z-score of the speed value gives the measure of dispersion of the from

the mean observed speed.

The probability that the speed of a car is between 63 km/h and 75 km/h is

0.273.

The given parameters are;

The mean of the speed of cars on the highway, $\overline x$ = 60 km/h

The standard deviation of the cars on the highway, σ = 5 km/h

Required:

The probability that the speed of a car is between 63 km/h and 75 km/h

Solution;

The z-score for a speed of 63 km/h is given as follows;

$Z=\dfrac{x-\bar x }{\sigma }$

Which gives;

$Z=\dfrac{63-60 }{5 } = 0.6$

From the z-score table, we have;

P(x < 63) = 0.7257

The z-score for a speed of 75 km/h is given as follows;

$Z=\dfrac{75-60 }{5 } = 3$

Which gives, P(x < 75) = 0.9987

The probability that the speed of a car is between 63 km/h and 75 km/h is therefore;

P(63 < x < 75) = P(x < 75) - P(x < 63) = 0.9987 - 0.7257 = 0.273

The probability that the speed of a car is between 63 km/h and 75 km/h is

0.273.

Learn more here:

brainly.com/question/17489087

7 0

2 years ago