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3 years ago

Simplify 1 · 0 - 0/y

1 answer:

3 0

The product of any number and 0 is zero. The expression is 1*0-0/y. 1*0 is obviously 0. 0/y is the same thing as 0 * 1/y, which is also zero. 0-0 is 0, so the answer is 0.

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Complete this activity. Interest at the rates of 5%, 6%, and 8% is earned on investments of $13,000, $12,000, and $4,000 respect

zvonat [6]

The answer is 1690. First find the amount of interest by a single investment by multiplying the investment amount by the percent interest in decimal form. Then calculate the total by adding up all the products of investment times interest. Matrices will allow to multiply two things together and add the products up all at the same time. So we need to find the investment matrix and interest matrix.

Investment matrix A (transposed) = (a1 a2 a3) = (13,000 12,000 4,000)

Interest Matrix B = (b1 b2 b3) = (0.05 0.06 0.08)

Amount of interest earned = a1b1 + a2b2 + a3b3

Amount of interest earned = (0.05) (13,000) + (0.06) (12,000) + (0.08)(4,000)

Amount of interest earned = 1690

7 0

4 years ago

I’d really appreciate help! I’m marking Brainliest.

Thepotemich [5.8K]

Answer

-6x+7y=11

Step-by-step explanation:

I don’t know if this is the right way

The slope of the line is y=6/7x+b

If you plug in the 2nd point

2=3/7+b

B=11/7

Y=6/7x+11/7

7y=6x+11

-6x+7y=11

8 0

3 years ago

N =64

faust18 [17]

Answer:

B. 1.70

Step-by-step explanation:

Given:

$n=64$
$\overline{x}=65.1$
$\sigma=24$
$\textsf{h}_0:\mu=60$
$\textsf{h}_1:\mu\neq 60$

$\textsf{If }\: X \sim\textsf{N}(\mu,\sigma^2)\:\textsf{ then }\: \overline{X} \sim \left(\mu,\dfrac{\sigma^2}{n}\right) \implies Z=\dfrac{\overline{X}-\mu}{\sigma / \sqrt{n}} \sim \textsf{N}(0,1)$

$\textsf{then test statistic is}: \quad z=\dfrac{\overline{x}-\mu}{\sigma / \sqrt{n}}$

Substituting the given values into the formula to find the test statistic z:

$\begin{aligned}\implies z &=\dfrac{65.1-60}{24 / \sqrt{64}}\\\\&=\dfrac{5.1}{3}\\\\&=\dfrac{17}{10}\\\\&=1.7\end{aligned}$

4 0

2 years ago

I need help on these 4 questions. Please someone help me. Don’t put a link. Please help me.

IrinaK [193]

Answer:

Median: b

Altitude: c

Angle bisector: a

Perpendicular bisector: d

Step-by-step explanation:

6 0

3 years ago

A vending machine sells chips at $0.40 and candy at $0.50. Last month, the vending machine yielded $139.80 with a sale of 298 it

Sidana [21]

Let us say... . the machine sold packages of chips at the amount of "p",
and the machine also sold, last month as well, an amount of "c" candies

now, last month the machine sold total 298 items
that yielded a revenue of $139.80

so...whatever "p" and "c" are, we know that, their sum for last month is,
298, so,
p + c = 298

how much revenue in sales of chips? well, they're 0.40 per package,
so, whatever "p" is, the machine sold 0.40 * p or 0.4p in revenue

how much for the candies? well, they're 0.50 for each, so
whatever "c" is, 0.50 * c or 0.5c in revenue

we know, their total revenue yield was 139.80
so
0.4p + 0.5c = 139.8

now... do substitution from the 1st equation on the 2nd one, and solve for "p", or packages of chips

that is $\begin{cases}
p+c=298\to c=\boxed{298-p}
\\ \quad \\
0.4p+0.5c=139.8
\\
\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\\
0.4p+0.5(\boxed{298-p})=139.8

\end{cases}\impliedby \textit{solve for "p"}$

6 0

3 years ago