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3 years ago

Please help me!!! :D

Mathematics

1 answer:

lesantik [10]3 years ago

4 0

Explanation:

For a theorem that says "if A then B", the converse is "if B then A."

The hing.e theorem has numerous parts to the hypothesis. Its converse retains many of those conditions, swapping only the relation between the included angle and the third side.

theorem: if two sides of one triangle are congruent to two sides of another, then the longest third side will be opposite the largest included angle.

converse: if two sides of one triangle are congruent to two sides of another, then the largest included angle will be opposite the longest third side.

Instead of relating the third side to the angle measure, the converse relates the angle measure to the third side.

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

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = the length of the calls, in minutes.

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = 0.1271

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05

P(Z > $\frac{x-4.5}{0.7}$ ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

$\frac{x-4.5}{0.7}=1.645$

${x-4.5}{}=1.645 \times 0.7$

x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0

4 years ago

Jennifer had 7/8 foot board she cut off 1/4 foot peice that was for a project.in feet how much board was left?

guapka [62]

$\frac{5}{8} \text{ feet board is left }$

Solution:

Given that, Jennifer had 7/8 foot board she cut off 1/4 foot peice that was for a project

To find: Board that was left

From given,

$\text{Total length of board } =\frac{7}{8} \text{ feet }$

$\text{Foot that was cut off } = \frac{1}{4} \text{ feet }$

Remaining feet of board is found by finding the difference between total length of board and foot that was cut off

$\text{Reamining board } = \text{Total length of board - feet of board that was cut off}\\$

Substituting the values we get,

$\text{Remaining board } = \frac{7}{8} - \frac{1}{4}\\\\\text{Remaining board } =\frac{7}{8} - \frac{1 \times 2}{4 \times 2}\\\\\text{Remaining board } =\frac{7}{8} -\frac{2}{8} = \frac{7-2}{8}\\\\\text{Remaining board } =\frac{5}{8}$

Thus $\frac{5}{8}$ feet board is left

5 0

3 years ago

Y+1 = 1/2 (x-6) solve for y in simplest form

Kruka [31]

Answer:

y = 1/2x -4

Step-by-step explanation:

Subtract 1 from both sides of the equation and simplify.

y = 1/2(x -6) -1

y = 1/2x -3 -1

y = 1/2x -4

3 0

3 years ago

Read 2 more answers

if there a 8 boys and 10 girls at a party what is the ratio of boys to girls A. 4 : 5 B. 5 : 4 C. 1 : 2

stellarik [79]

Answer:

A (4/5)

Step-by-step explanation:

Simplify to find the Answer:

1. 8/10 = 4/5!

6 0

3 years ago

Each CD is approximately 4.72 inches in circumference and 0.05 inches in depth. There are 15 CDs in the stack. Use the Cavalieri

Tju [1.3M]

Answer:

SA = 7.08 in²

Step-by-step explanation:

We can use the following method to solve the given problem

SA = 2B + LA

Where LA

LA = 4.72* 0.05* 15= 3.54

And

C = 2πr

Where

C = 4.72in

π = 3.14

Substituting, we have

4.72 = 2*3.14*r

Making r the subject of formula

r = 0.75in

Solve for the area of CD

A = πr²

Substituting the values we have

A = 3.14 *(0.75)²

A= 1.77 in²

Solving for surface area of the stack

SA = 2(1.77) + 3.54

SA = 7.08 in²

8 0

4 years ago