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3 years ago

Jenny has tossed a fair coin 25 times. It has landed on heads every single time. Is this possible? Why?

Mathematics

1 answer:

Cloud [144]3 years ago

7 0

Answer:

It is possible, but very unlikely.

Also, it is very likely the next flip will be Tails

This is because probability does not say if it is possible, but how likely it is!

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Consider the hypothesis test H0: μ1= μ2 against H1: μ1.Suppose that the sample sizes aren1 = 15 and n2 =μ2. Assume thatσ21 = σ22

torisob [31]

Answer:

see attachments

Step-by-step explanation:

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3 years ago

A mountain climber ascends 800 feet per hour from his original position. After 6 hours, his final position is 11,600 feet above

coldgirl [10]

<span>To find the beginning position, we need to know how many feet the climber has ascended in the 6-hour time frame. At the rate of 800 feet per hour, this would total to 4,800 feet in 6 hours (800 * 6). Subtracting this amount from the final position would give the elevation at the beginning of the ascent: (11600 - 4800) = 6,800 feet above sea level to begin.</span>

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4 years ago

What does it mean to cube something, thank you

nadezda [96]

To multiply it by itself three time eg 2x2x2

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3 years ago

Read 2 more answers

Hya think of a number "m" . She square it, add "y" to the answer and then subtract 7 from it. The final answer is "A". 1) Derive

Bond [772]

<h2><u>Solution (1)</u> :</h2>

Given, to find A we have to :

square m
Add y to m²
Subtract 7 from m² + y

From the question, the following equation can be formed :

$=\tt A = {m}^{2} + y - 7$

Therefore, the formula for finding A = m² + y - 7

<h2><u>Solution (2)</u> :</h2>

The value of A we can derive from the formula is :

$=\tt A = {m}^{2} + y - 7$

Value of m = 3 (given)

Which means :

$=\tt A = {3}^{2} + y - 7$

$=\tt \: A = 9 + y - 7$

$=\tt A = 9 - 7 + y$

$\color{plum} =\tt A = 2 + y$

Thus, the value of A = 2+y

Therefore, the value of A = <u>2+y</u>

7 0

3 years ago

Can someone please help me on number 16-ABC

melomori [17]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the inequality

-2x < 10

-6 < -2x

<u>Part a) Is x = 0 a solution to both inequalities</u>

FOR -2x < 10

substituting x = 0 in -2x < 10

-2x < 10

-3(0) < 10

0 < 10

TRUE!

Thus, x = 0 satisfies the inequality -2x < 10.

∴ x = 0 is the solution to the inequality -2x < 10.

FOR -6 < -2x

substituting x = 0 in -6 < -2x

-6 < -2x

-6 < -2(0)

-6 < 0

TRUE!

Thus, x = 0 satisfies the inequality -6 < -2x

∴ x = 0 is the solution to the inequality -6 < -2x

Conclusion:

x = 0 is a solution to both inequalites.

<u>Part b) Is x = 4 a solution to both inequalities</u>

FOR -2x < 10

substituting x = 4 in -2x < 10

-2x < 10

-3(4) < 10

-12 < 10

TRUE!

Thus, x = 4 satisfies the inequality -2x < 10.

∴ x = 4 is the solution to the inequality -2x < 10.

FOR -6 < -2x

substituting x = 4 in -6 < -2x

-6 < -2x

-6 < -2(4)

-6 < -8

FALSE!

Thus, x = 4 does not satisfiy the inequality -6 < -2x

∴ x = 4 is the NOT a solution to the inequality -6 < -2x.

Conclusion:

x = 4 is NOT a solution to both inequalites.

Part c) Find another value of x that is a solution to both inequalities.

<u>solving -2x < 10</u>

$-2x\:$

Multiply both sides by -1 (reverses the inequality)

$\left(-2x\right)\left(-1\right)>10\left(-1\right)$

Simplify

$2x>-10$

Divide both sides by 2

$\frac{2x}{2}>\frac{-10}{2}$

$x>-5$

$-2x-5\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-5,\:\infty \:\right)\end{bmatrix}$

<u>solving -6 < -2x</u>

-6 < -2x

switch sides

$-2x>-6$

Multiply both sides by -1 (reverses the inequality)

$\left(-2x\right)\left(-1\right)$

Simplify

$2x$

Divide both sides by 2

$\frac{2x}{2}$

$x$

$-6$

Thus, the two intervals:

$\left(-\infty \:,\:3\right)$

$\left(-5,\:\infty \:\right)$

The intersection of these two intervals would be the solution to both inequalities.

$\left(-\infty \:,\:3\right)$ and $\left(-5,\:\infty \:\right)$

As x = 1 is included in both intervals.

so x = 1 would be another solution common to both inequalities.

<h3>SUBSTITUTING x = 1</h3>

FOR -2x < 10

substituting x = 1 in -2x < 10

-2x < 10

-3(1) < 10

-3 < 10

TRUE!

Thus, x = 1 satisfies the inequality -2x < 10.

∴ x = 1 is the solution to the inequality -2x < 10.

FOR -6 < -2x

substituting x = 1 in -6 < -2x

-6 < -2x

-6 < -2(1)

-6 < -2

TRUE!

Thus, x = 1 satisfies the inequality -6 < -2x

∴ x = 1 is the solution to the inequality -6 < -2x.

Conclusion:

x = 1 is a solution common to both inequalites.

7 0

3 years ago