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3 years ago

I need help with Standard Deviation!

1 answer:

3 0

A: The middle 95% contain between 23.8 and 24.2 oz.(2 SDs under and over the mean) B: Approximately 68% of boxes have between 23.9 and 24.1 oz of cereal (1 SD under and over the mean) C: 2.5% of boxes contain more than 24.2, because 95% are within 2 SDs, and that includes lower and higher extremes. D: 16% of boxes have more than 24.1, because 68% are within 1 SD of the mean, so 32 are left, and that includes lower and higher extremes.

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HELP PLEASE I NEED HELP

attashe74 [19]

Answer:

I think it is 370

Step-by-step explanation:

5 0

3 years ago

The sale price of an item on sale for 40% off is $48. What was the regular price?

Charra [1.4K]

Good evening ,

Answer:

<h2>the regular price of the item is $80.</h2>

Step-by-step explanation:

Let p represent the regular price of the item.

p - p×(40/100) = 48 ⇌ p×(1 - 40/100) = 48 ⇌ p×(100/100 - 40/100) = 48

⇌ p×(60/100) = 48 ⇌ p = (480×100)/60 = 48 000/60 = 80.

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5 0

3 years ago

Read 2 more answers

is 9/9 a whole number? because when simplified it equls 1 and 1 is a whole number, but since its in fractiom form does that coun

kifflom [539]

Answer:

Even though it's in fraction form it's still a whole number.

Step-by-step explanation:

Here's an example to make this easier-

If you cut an orange into 8 slices and count all the parts it's 8/8, but you still have 1 whole orange. I hope this makes sense! have a great day! ^^

8 0

1 year ago

What reasoning and explanations can be used when solving radical equations and how do extraneous solutions arise from radical eq

Mkey [24]

Answer:

We know that, a 'radical equation' is an equation that contains radical expressions, which further are the expressions containing radicals ( square roots and other roots of numbers ).

In order to solve radical equations, we use the rules of exponents and basic algebraic properties.

The common reasoning to use while solving a radical equation is:

1. Isolate the radical expression.

2. Square both sides of the equation to remove radical.

3. After removing the radical, solve the equation to find the unkown variable

4. Check the answer for the errors occurred by removing the radicals.

For e.g. $\sqrt{x}-3=5$

i.e. $\sqrt{x}-3+3=5+3$ ( Adding 3 on both sides )

i.e. $\sqrt{x}=8$

i.e. $(\sqrt{x})^{2} =8^{2}$

i.e. $x=64$ .

So, the solution the the radical equation $\sqrt{x}-3=5$ is x = 64.

Further, we know that an 'extraneous solution' is that root of the radical equation which is not a root of the original equation and is excluded from the domain.

for e.g. Take $\sqrt{x+4} =x-2$