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gladu [14]
3 years ago
5

LOTS OF POINTS!!! HELP ASAP Consider triangle QRS. The legs each have a length of 10 units. What is the length of the hypotenuse

of the triangle? 5 units 5 square root of 2 units 10 units 10 square root of 2 units

Mathematics
2 answers:
saveliy_v [14]3 years ago
8 0

Answer:

10\sqrt{2}

Step-by-step explanation:

Pythagorean Theorem

QS^2+SR^2=QR^2

10^2+10^2=QR^2

100+100=QR^2

200=QR^2

\sqrt{200}=QR

10\sqrt{2}

AysviL [449]3 years ago
3 0

Answer:

In right Δ QRS, right angled at , S.

By Pythagorean theorem

→(Hypotenuse)²=(Altitude)^2 +(Base)²

→ Q S² + SR² = QR²

→10² +10²= QR²

→ QR²=100+100

→ QR²=200

QR=\sqrt{200}\\\\ QR=\sqrt{10 \times 10 \times 2}\\\\ QR=10\sqrt{2}

Option D

Hypotenuse=10\sqrt{2} units

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Heights of men on a baseball team have a​ bell-shaped distribution with a mean of and a standard deviation of . Using the empiri
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Let assume that the mean is 184 and the standard deviation is 6

Heights of men on a baseball team have a​ bell-shaped distribution with a mean 184 of and a standard deviation of 6 . Using the empirical​ rule, what is the approximate percentage of the men between the following​ values? a.166 cm and 202 cm b. 172 cm and 196cm

Answer:

P(156<X<202) =  99.7%

P(172<X<196) =  95.5%

Step-by-step explanation:

Given that :

Heights of men on a baseball team have a​ bell-shaped distribution with a mean of and a standard deviation of . Using the empirical​ rule, what is the approximate percentage of the men between the following​ values? a.166 cm and 202 cm b. 172 cm and 196cm

For a.

Using the empirical​ rule, what is the approximate percentage of the men between the following​ values 166 cm and 202 cm.

the z score can be determined by using the formula:

z = \dfrac{X - \mu}{\sigma}

z(166) = \dfrac{166-184}{6}

z(166) = \dfrac{-18}{6}

z(166) = -3

z(202) = \dfrac{202-184}{6}

z(202) = \dfrac{18}{6}

z(202) = 3

P(156<X<202) = P( μ - 3σ < X < μ + 3σ )

P(156<X<202) = P( - 3 < Z < 3)

P(156<X<202) = P( Z <  3) - P(Z < -3)

P(156<X<202) = 0.99865-  0.001349

P(156<X<202) =  0.997301

P(156<X<202) =  99.7%

For b.

b. 172 cm and 196cm

z = \dfrac{X - \mu}{\sigma}

z(172) = \dfrac{172-184}{6}

z(172) = \dfrac{-12}{6}

z(172) = -2

z(196) = \dfrac{196-184}{6}

z(196) = \dfrac{12}{6}

z(196) = 2

P(172<X<196) = P( μ - 2σ < X < μ + 2σ )

P(172<X<196) = P( - 2 < Z < 2)

P(172<X<196) = P( Z <  2) - P(Z < -2)

P(172<X<196) = 0.9772 -  0.02275

P(172<X<196) =  0.95445

P(172<X<196) =  95.5%

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Step-by-step explanation and answer:

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Answer:

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Step-by-step explanation:

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