What is the simplified expression for 6 to the power of 4 multiplied by 6 to the power of 3 over 6 to the power of 5?

H + 732 = -194 <br> Solve for H

lilavasa [31]

H + 732 = -194

* combine like terms

H = -194 - 732

H = -926

4 0

3 years ago

What is the diameter of a human hair written in scientific notation?

Ymorist [56]

diameter of human hair is around 0.0025cm

in scientific notation it is: 2.5x10^-3

5 0

3 years ago

What is the sum of the first 6 terms of this geometric sequence? –4, –16, –64, –256, …

MatroZZZ [7]

$a_1=-4;\ a_2=-16;\ a_3=-64;\ a_4=-256;\ ...\\\\r=\dfrac{a_{n+1}}{a_n}\to r=\dfrac{a_2}{a_1}\to r=\dfrac{-16}{-4}=4$

The formula of the sum of a geometric sequence:

$S_n=a_1\cdot\dfrac{1-r^n}{1-r}$

We have:

$a_1=-4;\ n=6;\ r=4$

substitute:

$S_6=-4\cdot\dfrac{1-4^6}{1-4}=-4\cdot\dfrac{1-4096}{-3}=-4\cdot\dfrac{-4095}{-3}=-4\cdot1365=-5460$

Answer: C. -5460

6 0

4 years ago

prescription for marilyn jones piroxicam 20 mg capsule take 1 capsule 2 times a day how many milligrams mg will marilyn take of

guapka [62]

Answer:

she will be taking 3600 mgs. in 90 days

Step-by-step explanation:

if she is taking a 20 mgs. twice a day that will make it 40 mgs. a day. If she takes it for 90 days

you will multiply 90 by 40

so

90

x40

= 3600 mgs. in 90 days

3 0

3 years ago

Read 2 more answers

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

7 0

3 years ago