Ask question

Ask question

All categories

murzikaleks [220]

3 years ago

6

Which scenario involves the value of greatest magnitude?

1 answer:

Sonbull [250]3 years ago

5 0

Answer:

D. A withdrawal of $30 from a bank account

Step-by-step explanation:

When it says magnitude it doesn't matter if it's a positive or negative number.

So it's just asking which number is the greatest which is 30.

I hope my explanation makes sense and helps you out.

You might be interested in

Identify the expression that is equivalent to cot x/ csc x -1

Yuri [45]

Cotx = cosx/sinx
cscx = 1/sinx

If we plug these into your equation we get:
(cosx/sinx)/(1/sinx) - 1

Dividing by a number is the same as multiplying by its reciprocal so it can be rewritten as:
(cosx/sinx)(sinx) - 1

The two sinx's cancel each other out so we end up with cosx - 1

8 0

3 years ago

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Amanda [17]

Answer:

A man is driving to pick his daughter from school. He waits for his daughter to be done, then he drives back home.

7 0

3 years ago

I will give brainliest please hurry

Eduardwww [97]

Answer:

B)

Step-by-step explanation:

300, since the line inside the box is on 300

Hope that helps!

6 0

3 years ago

Read 2 more answers

Mr. Duncan bought a table at a discount if 30% thus saving $42. what was the marked price of the table?

katovenus [111]

Answer:

$140

Step-by-step explanation:

if 30% of full price = $42 then 100% = 42/0.3 = 140

5 0

2 years ago

Read 2 more answers

Suppose a simple random sample of size nequals36 is obtained from a population with mu equals 74 and sigma equals 6. (a) Descri

OlgaM077 [116]

Part a)

The simple random sample of size n=36 is obtained from a population with

$\mu = 74$

and

$\sigma = 6$

The sampling distribution of the sample means has a mean that is equal to mean of the population the sample has been drawn from.

Therefore the sampling distribution has a mean of

$\mu = 74$

The standard error of the means becomes the standard deviation of the sampling distribution.

$\sigma_ { \bar X } = \frac{ \sigma}{ \sqrt{n} } \\ \sigma_ { \bar X } = \frac{ 6}{ \sqrt{36} } = 1$

Part b) We want to find

$P(\bar X \:>\:75.9)$

We need to convert to z-score.

$P(\bar X \:>\:75.9) = P(z \:>\: \frac{75.9 - 74}{1} ) \\ = P(z \:>\: \frac{75.9 - 74}{1} ) \\ = P(z \:>\: 1.9) \\ = 0.0287$

Part c)

We want to find

$P(\bar X \: < \:71.95)$

We convert to z-score and use the normal distribution table to find the corresponding area.

$P(\bar X \: < \:71.95) = P(z \: < \: \frac{71.9 5- 74}{1} ) \\ = P(z \: < \: \frac{71.9 5- 74}{1} ) \\ = P(z \: < \: - 2.05) \\ = 0.0202$

Part d)

We want to find :

$P(73\:$

We convert to z-scores and again use the standard normal distribution table.

$P( \frac{73 - 74}{1} \:< \: z$

5 0

3 years ago