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2 years ago

6

A coin is tossed four times. What is the probability of getting tails four times in a row?

1 answer:

Anvisha [2.4K]2 years ago

7 0

Answer:

b) 1/32

Step-by-step explanation:

1/2 x 1/2 x 1/2 x 1/2 = 1/32

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Solve: Vx+1= Vx+6 -1

swat32

Answer:

$x + 1 = x +7-2\sqrt{x+6}$

Step-by-step explanation:

$\sqrt{x+1}= \sqrt{x+6}-1$

Squaring both sides of the equation would give $(\sqrt{x+1} )^{2} =(\sqrt{x+6}-1 )^{2}$ which then simplifies to $x + 1 = x -2\sqrt{x+6}+7$

We can then subtract 7 from both sides to get $x-6=x-2\sqrt{x+6}$

Then we can subtract x from both sides, which cancel out and we get $-6 =-2\sqrt{x+6}$

Then we need to divide both sides by -2 to get $3=\sqrt{x+6}$

Then we need to square both sides $(3)^{2} =(\sqrt{x+6} )^{2}$ to get rid of the square root to get 9 = x + 6

Then finally subtracting 6 from both sides we get 3 = x

3 0

3 years ago

4. What percent of all computers sold during the entire week<br><br> were sold on Friday!

UkoKoshka [18]

Answer:

I confused... there are no numbers

7 0

3 years ago

A shirt originally cost $52.29, but it is on sale for $36.60. What is the percentage decrease of the price of the shirt? If nece

Fantom [35]

Answer: 30%

Step-by-step explanation:

6 0

3 years ago

How do i solve this equation 13-4x=1-x ?

Rainbow [258]

<span>go</span> to symbolad dot com

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3 years ago

Read 2 more answers

Average rate of change help!!

frez [133]

Given that the function is $f(x)=x^{3}-6 x^{2}+4 x+7$

We need to determine the average rate of change over the interval $0 \leq x \leq 5$

<u>Value of f(x) when x = 0:</u>

Substituting x = 0 in the function $f(x)=x^{3}-6 x^{2}+4 x+7$ , we have;

$f(0)=(0)^{3}-6 (0)^{2}+4 (0)+7$

$f(0)=0-0+0+7$

$f(0)=7$

Thus, the value of f(0) is 7.

<u>Value of f(x) when x = 5:</u>

Substituting x = 5 in the function $f(x)=x^{3}-6 x^{2}+4 x+7$ , we have;

$f(5)=(5)^{3}-6 (5)^{2}+4 (5)+7$

$f(5)=125-150+20+7$

$f(5)=2$

Thus, the value of f(5) is 2.

<u>Average rate of change:</u>

The average rate of change can be determined using the formula,

$Rate \ of \ change=\frac{f(b)-f(a)}{b-a}$

where $a=0$ and $b=5$

Thus, we have;

$Rate \ of \ change=\frac{f(5)-f(0)}{5-0}$

$Rate \ of \ change=\frac{2-7}{5-0}$

$Rate \ of \ change=\frac{-5}{5}$

$Rate \ of \ change=-1$

Thus, the average rate of change over the interval $0 \leq x \leq 5$ is -1.

3 0

3 years ago