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2 years ago

Find the exact solution to the equation. 320 (1/4 )^ x/4 = 5

1 answer:

vladimir2022 [97]2 years ago

8 0

You have to create and equivalent EXPRESSION in the equation that have equal bases then solve for x
SO THE ANSWER IS 12

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I’m another normally distributed test score is 80 and the standard deviations is 10. How common are values in the less than 60?

lord [1]

Answer:

The probability of obtaining a score less than 60 is 2.28%

It happens 2 out of 100 times.

Step-by-step explanation:

If the mean μ = 80 and the standard deviation σ = 10, then we need to find the probability that an X value is less than 60.

Then we find

$P(X$

To find this probability we use the Z statistic.

$Z = \frac{X- \mu}{\sigma}$

$P(\frac{X- \mu}{\sigma}$

$P(Z$

This is the same as

$P(Z> 2)$

We look for this value in the table for the normal distribution of right queue and we have:

$P(X 2) = 0.02275$

The probability of obtaining a score less than 60 is 2.28%

It happens 2 out of 100 times.

6 0

2 years ago

What is this, x-8 all over 3=3

inysia [295]

You mean (x-8)/3 = 3 ...right?

You would multiply both sides by 3.
Then, add 8 to both sides.

x=17

5 0

2 years ago

What is the sum?

BaLLatris [955]

4(x + 6)
4 × x = 4x
4 × 6 = 24

3(3x + 4)
3 × 3x = 9x
3 × 4 = 12

9x + 4x = 13x
12 + 24 = 36

So your answer is B) 13x + 36. I hope this helps!

4 0

3 years ago

Read 2 more answers

If 9 x 0 + 4 = 4 had parenthesis where would they go?

Bingel [31]

Answer:

(9 x 0) + 4=4

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Find all values of c such that c/(c - 5) = 4/(c - 4). If you find more than one solution, then list the solutions you find separ

Deffense [45]

Answer:

$\large\boxed{\bold{NO\ REAL\ SOLUTIONS}}\\\boxed{x=4-2i,\ x=4+2i}$

Step-by-step explanation:

$Domain:\\c-4\neq0\ \wedge\ c-5\neq0\Rightarrow c\neq4\ \wedge\ c\neq5\\\\\dfrac{c}{c-5}=\dfrac{4}{c-4}\qquad\text{cross multiply}\\\\c(c-4)=4(c-5)\qquad\text{use the distributive property}\\\\c^2-4c=4c-20\qquad\text{subtract}\ 4c\ \text{from both sides}\\\\c^2-8c=-20\qquad\text{add 20 to both sides}\\\\c^2-8c+20=0\qquad\text{use the quadratic formula}$

$\text{for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac0,\ \text{then the equation has two solutions}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x^2-8x+20=0\\\\a=1,\ b=-8,\ c=20\\\\b^2-4ac=(-8)^2-4(1)(20)=64-80=-16$

$\text{In the set of complex numbers:}\\\\i=\sqrt{-1}\\\\\text{therefore}\ \sqrt{b^2-4ac}=\sqrt{-16}=\sqrt{(16)(-1)}=\sqrt{16}\cdot\sqrt{-1}=4i\\\\x=\dfrac{-(-8)\pm4i}{2(1)}=\dfrac{8\pm4i}{2}=\dfrac{8}{2}\pm\dfrac{4i}{2}=4\pm2i$

6 0

2 years ago

Read 2 more answers