Ask question

Ask question

All categories

2 years ago

15

What is the solution to the equation 1/square root of 8 = 4^(m + 2)?

2 answers:

Digiron [165]2 years ago

8 0

Take the log (base 4) of both sides of the equation.
$\log_{4}(\frac{1}{\sqrt{8}}) = m + 2$
$\frac{-1}{2} \log_{4}(4^{\frac{3}{2}})-2 = m$
$-2 \frac{3}{4}= m$

The appropriate choice is ...
m = -11/4

salantis [7]2 years ago

6 0

I agree with the other person, the answer is -11/4

You might be interested in

Jessie’s friend Paul runs a hot dog stand near the car plant. Jessie gets 3 hot dogs and pays Paul $3.60

Deffense [45]

3.60 divided by 3 = h

3 0

2 years ago

Read 2 more answers

What is the nearest value of d to the nearest hundredth

Lisa [10]

The Answer to question is 2.05

6 0

2 years ago

Read 2 more answers

To save money, a local charity organization wants to target its mailing requests for donations to individuals who are most suppo

mars1129 [50]

Answer:

The 80% confidence interval for difference between two means is (0.85, 1.55).

Step-by-step explanation:

The (1 - <em>α</em>) % confidence interval for difference between two means is:

$CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2,(n_{1}+n_{2}-2)}\times SE_{\bar x_{1}-\bar x_{2}}$

Given:

$\bar x_{1}=M_{1}=6.1\\\bar x_{2}=M_{2}=4.9\\SE_{\bar x_{1}-\bar x_{2}}=0.25$

Confidence level = 80%

$t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.20/2, (5+5-2)}=t_{0.10,8}=1.397$

*Use a <em>t</em>-table for the critical value.

Compute the 80% confidence interval for difference between two means as follows:

$CI=(6.1-4.9)\pm 1.397\times 0.25\\=1.2\pm 0.34925\\=(0.85075, 1.54925)\\\approx(0.85, 1.55)$

Thus, the 80% confidence interval for difference between two means is (0.85, 1.55).

3 0

2 years ago

Hey im in online class RN can does anyone know what (-4) – 2 = please respond early

rjkz [21]

Answer:

(-4) – 2 = -6

Hope This Helps.!! :)

5 0

2 years ago

Can someone solve this?

ladessa [460]

First question:

Recall that $\cos^2x+\sin^2x=1$ and $\sqrt{x^2}=|x|$ for all $x$ . So

$\sqrt{1-\cos^2x}=\sqrt{\sin^2x}=|\sin x|$

$\sqrt{1-\sin^2x}=\sqrt{\cos^2x}=|\cos x|$

For $0$ , we expect both $\cos x>0$ and $\sin x>0$ (i.e. the sine and cosine of any angle that lies in the first quadrant must be positive). By definition of absolute value, $|x|=x$ if $x>0$ .

So we have

$\dfrac{\sqrt{1-\cos^2x}}{\sin x}+\dfrac{\sqrt{1-\sin^2x}}{\cos x}=\dfrac{\sin x}{\sin x}+\dfrac{\cos x}{\cos x}=1+1=\boxed{2}$

making H the answer.

Second question:

C is always true, because the inequality reduces to x > y.

6 0

2 years ago