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3 years ago

Find the value of x in this figure.

Mathematics

2 answers:

kondor19780726 [428]3 years ago

6 0

ANSWER

The value of x is 47°

EXPLANATION

PQ is a tangent to the circle at Q.

This tangent meets the diameter at 90°.

The sum of interior angles of a triangle is 180°

This implies that:

$90 \degree + x + 43 \degree = 180 \degree$

$133 \degree + x = 180 \degree$

Group similar terms to obtain;

$x = 180 \degree - 133 \degree$

Simplify similar terms to get;

$x = 47\degree$

Rudiy273 years ago

6 0

Answer:

The value of x = 47°

Step-by-step explanation:

From the figure we can see that a circle with center O.

PQ is a tangent to the circle fro point P.

m<P = 43°

Therefore <Q = 90°

<u>To find the value of x</u>

From the given triangle we can write,

x + m<Q + m<P = 180

x = 180 - (m<Q + m<P)

= 180 - (90 + 43)

= 180 - 133 = 47°

Therefore the value of x = 47°

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The value of x when y is 8 is 9.75.

It is given that

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The constant of proportionality is the ratio that relates two given values.

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3 years ago

The entire league division that the Hawks belong to has the same ratio of red heads to everyone else. What is the total number o

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This question is incomplete

Complete Question

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3 years ago

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Anna [14]

Answer:

a) 0.4452

b) 0.0548

c) 0.0501

d) 0.9145

e) 6.08 minutes or greater

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.7 minutes

Standard Deviation, σ = 0.50 minutes.

We are given that the distribution of length of the calls is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(calls last between 4.7 and 5.5 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{4.7 - 4.7}{0.50} \leq z \leq \displaystyle\frac{5.5-4.7}{0.50}) = P(0 \leq z \leq 1.6)\\\\= P(z \leq 1.6) - P(z$

$P(4.7 \leq x \leq 5.5) = 44.52\%$

b) P(calls last more than 5.5 minutes)

$P(x > 5.5) = P(z > \displaystyle\frac{5.5-4.7}{0.50}) = P(z > 1.6)\\\\P( z > 1.6) = 1 - P(z \leq 1.6)$

Calculating the value from the standard normal table we have,

$1 - 0.9452 = 0.0548 = 5.48\%\\P( x > 5.5) = 5.48\%$

c) P( calls last between 5.5 and 6 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{5.5 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(1.6 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$

$P(5.5 \leq x \leq 6) = 5.01\%$

d) P( calls last between 4 and 6 minutes)

$P(4 \leq x \leq 6) = P(\displaystyle\frac{4 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(-1.4 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$