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3 years ago

Can a right triangle have a leg that is 10 meters long and a hypotenuse that is 10 meters long

2 answers:

5 0

Lets see
a and b are legs
c=hypotnuse
a^2+b^2=c^2
10^2+b^2=10^2
100+b^2=100
minus 100 both sides
b^2=0
b=0
false, if you have a leg legnth 0, then they lines are right on top of each other

no cannot

bogdanovich [222]3 years ago

3 0

Solutions

To solve this problem we have to use the Pythagorean theorem. You can only use the Pythagorean theorem in Right Triangles. The longest side of the triangle is called the "hypotenuse". C² is the longest side so it is the hypotenuse . To calculate c² we have to do α² + β² = c².

Given

α² = 10
c² = 10

Solve

α² + β² = c²
α² (10²) + β² = c² (10²)
100 + β² = 100
β² = 100 - 100 = 0

Simplify

A right triangle can not have a leg that is 10 meters long and a hypotenuse that is 10 meters long.

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3 years ago

A university wants to compare out-of-state applicants' mean SAT math scores (?1) to in-state applicants' mean SAT math scores (?

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Answer:

d. Yes, because the confidence interval does not contain zero.

Step-by-step explanation:

We are given that the university looks at 35 in-state applicants and 35 out-of-state applicants. The mean SAT math score for in-state applicants was 540, with a standard deviation of 20.

The mean SAT math score for out-of-state applicants was 555, with a standard deviation of 25.

Firstly, the Pivotal quantity for 95% confidence interval for the difference between the population means is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean SAT math score for in-state applicants = 540

$\bar X_2$ = sample mean SAT math score for out-of-state applicants = 555

$s_1$ = sample standard deviation for in-state applicants = 20

$s_2$ = sample standard deviation for out-of-state applicants = 25

$n_1$ = sample of in-state applicants = 35

$n_2$ = sample of out-of-state applicants = 35

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(35-1)\times 20^{2} +(35-1)\times 25^{2} }{35+35-2} }$ = 22.64

Here for constructing 95% confidence interval we have used Two-sample t test statistics.

So, 95% confidence interval for the difference between population means ( $\mu_1-\mu_2$ ) is ;

P(-1.997 < $t_6_8$ < 1.997) = 0.95 {As the critical value of t at 68 degree

of freedom are -1.997 & 1.997 with P = 2.5%}

P(-1.997 < $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 1.997) = 0.95

P( $-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ${(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}$ < $1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.95

P( $(\bar X_1-\bar X_2)-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.95

95% confidence interval for ( $\mu_1-\mu_2$ ) =

[ $(\bar X_1-\bar X_2)-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ]

=[ $(540-555)-1.997 \times {22.64 \times \sqrt{\frac{1}{35} +\frac{1}{35} } }$ , $(540-555)+1.997 \times {22.64 \times \sqrt{\frac{1}{35} +\frac{1}{35} } }$ ]

= [-25.81 , -4.19]

Therefore, 95% confidence interval for the difference between population means SAT math score for in-state and out-of-state applicants is [-25.81 , -4.19].

This means that the mean SAT math scores for in-state students and out-of-state students differ because the confidence interval does not contain zero.

So, option d is correct as Yes, because the confidence interval does not contain zero.

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3 years ago

If Ryan paid $63 for shoes that were regularly priced at $90, what was the percent discount of the

Ray Of Light [21]

Answer:

30% off

Step-by-step explanation:

63 is 70% of 90 which means that 30 percent of the price is off.

Please mark me brainliest. I need it. Hope this helps

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