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2 years ago

Can somebody help me with this question ;-;

Mathematics

1 answer:

Hatshy [7]2 years ago

6 0

Answer:

I believe the answer is 5 units. Because it shows the line's arrow at unit 5.

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What is the ratio of the sides of triangle XYZ? --> can anyone solve this equation ?

lubasha [3.4K]

Answer:

1:2:√3

Step-by-step explanation:

since XYZ is a special right triangle

8 0

2 years ago

What are the solutions to x^2-5x+3=0?

Viktor [21]

Answer:

Step-by-step explanation:

You need to use the quadratic equation

x = $\frac{-b +/- \sqrt{b^2 - 4ac} }{2a}$

Givens

a = 1
b = - 5
c = 3

x = $\frac{5 +/- \sqrt{5^2 - 4a*1*3} }{2} \\$

x =( 5 +/- sqrt(25 - 12) ) / 2

x = (5 +/- sqrt(13) )/2

x = (5 + sqrt(13) / 2

x = 4.303 rounded

x = (5 - sqrt(13) ) /2

x = .6972

4 0

3 years ago

Graph your linear equation. y=x÷3

mojhsa [17]

Here ya go
(correct me if wrong)

7 0

2 years ago

2. Reece is playing a carnival game in which he

Maslowich

25% or 2/4 and to win one game 50% or 1/2

6 0

2 years ago

Construct a 99% confidence interval for the population mean, mu. Assume the population has a normal distribution. A group of 1

Zarrin [17]

Answer:

99% confidence interval for the population mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% 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 age of selected students = 22.4 years

s = sample standard deviation = 3.8 years

n = sample of students = 19

$\mu$ = population mean

Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.878 < $t_1_8$ < 2.878) = 0.99 {As the critical value of t at 18 degree of

freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.878) = 0.99

P( $-2.878 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.878 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ , $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ]

= [ $22.4 -2.878 \times {\frac{3.8}{\sqrt{19} }$ , $22.4 +2.878 \times {\frac{3.8}{\sqrt{19} }$ ]

= [19.891 , 24.909]

Therefore, 99% confidence interval for the population mean is [19.891 , 24.909].

6 0

3 years ago

Can somebody help me with this question ;-;​

Can somebody help me with this question ;-;