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3 years ago

14

Help please I don't get it !!!!!!!! :(((((

1 answer:

anygoal [31]3 years ago

7 0

Answers is C . check it out

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Results for: Timothy owns a t shirt printing buisnesses the expenses are $4 per t shirt Timothy sales each shirt for $10 he sold

topjm [15]

You have to do subtract 10 from 4 because 6 is how much we will profit from each shirt. Now, 6 times 327 is $1,962. That's how much Timothy earned of profit.

8 0

3 years ago

What is equivalent to the square root of -27

lions [1.4K]

For the square root of a negative number, you would have to get into imaginary numbers. If you haven't learned about it yet, then the answer would be no real solutions. But if you have, here's how to solve it:

First, identify that 9 x 3 = 27. To after taking the square root, you will get 3√3 (Since the square root of 9 is 3 and there's no square root of 3, so leave it in the √)

However, since it is a negative number, you will have to include the letter i in the answer. i represents the imaginary part of it since you can't have a negative number in the square root.

So your answer will look like 3i√3.

3 0

3 years ago

What is the component form of the vector represented by 2a when a⃗ =〈−2,3〉?

Paladinen [302]

Answer:

$\vec v = 2\cdot \vec a = \langle -4,6 \rangle$

Step-by-step explanation:

Let $\vec a = \langle -2, 3 \rangle$ , we need to determine $\vec v = 2\cdot \vec a$ , which can be found by applying the following property:

$\alpha \cdot \langle x,y \rangle = \langle \alpha \cdot x, \alpha \cdot y\rangle$ , $\forall \,\alpha,x,y\in \mathbb{R}$ (1)

Then,

$\vec v = 2\cdot \langle -2,3 \rangle$

$\vec v = \langle -4,6 \rangle$

8 0

3 years ago

Amy's salary is 40,000 in her first year on the job with a 2,000 rise every year there for

Elena-2011 [213]

Answer:

F

Step-by-step explanation:

8 0

3 years ago

A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73

faust18 [17]

Answer:

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher

Step-by-step explanation:

Data provided in the question:

sample size, n = 103

Mean temperature, μ = 98.3

°

Standard deviation, σ = 0.73

Degrees of freedom, df = n - 1 = 102

Now,

For Confidence level of 99%, and df = 102, the t-value = 2.62 [from the standard t table]

Therefore,

CI = $(Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})$

Thus,

Lower limit of CI = $(Mean - \frac{t\times\sigma}{\sqrt{n}})$

or

Lower limit of CI = $(98.3 - \frac{2.62\times0.73}{\sqrt{103}})$

or

Lower limit of CI = 98.11

and,

Upper limit of CI = $(Mean + \frac{t\times\sigma}{\sqrt{n}})$

or

Upper limit of CI = $(98.3 + \frac{2.62\times0.73}{\sqrt{103}})$

or

Upper limit of CI = 98.49

Hence,

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher and the mean temperature could very possibly be 98.6°F

7 0

3 years ago