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soldier1979 [14.2K]
3 years ago
10

Which of these sets of ordered pairs are functions?

Mathematics
1 answer:
Arisa [49]3 years ago
4 0

Answer: The second one if im wrong please dont get maddd :(

but 80% im ure its right

Step-by-step explanation:

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2.
makkiz [27]

Answer:2,695

Step-by-step explanation: just add.

3 0
2 years ago
I need help with this
borishaifa [10]

Answer:

  ★ = 3

  ✚ = 4

  ■ = 4

  ◆ = 5

Step-by-step explanation:

If we use direct correspondence in the second equation, then we have ...

  ◆ = 5

  ★ = 3

The values of ★ and ✚ in the first equation must total 7. This means ...

  ✚ = 7 -3 = 4

Again, using direct correspondence in the third equation, we have ...

  ■ = 4

6 0
3 years ago
From a random sample of 58 businesses, it is found that the mean time the owner spends on administrative issues each week is 21.
zzz [600]

Answer: (20.86, 22.52)

Step-by-step explanation:

Formula to find the confidence interval for population mean :-

\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}

, where \overline{x} = sample mean.

z*= critical z-value

n= sample size.

\sigma = Population standard deviation.

By considering the given question , we have

\overline{x}= 21.69

\sigma=3.23

n= 58

Using z-table, the critical z-value for 95% confidence = z* = 1.96

Then, 95% confidence interval for the amount of time spent on administrative issues will be :

21.69\pm (1.96)\dfrac{3.23}{\sqrt{58}}

=21.69\pm (1.96)\dfrac{1.7}{7.61577}

=21.69\pm (1.96)(0.223221)

\approx21.69\pm0.83

=(21.69-0.83,\ 21.69+0.83)=(20.86,\ 22.52)

Hence, the 95% confidence interval for the amount of time spent on administrative issues = (20.86, 22.52)

7 0
3 years ago
Find an equation in standard form for the hyperbola with vertices at (0,plus or minus 6) and asymptotes at y= (plus or minus 3/5
antoniya [11.8K]

Answer:

\frac{(y)^2}{36}-\frac{(x)^2}{100}=1

Step-by-step explanation:

Given:

Vertices of Hyperbola : (0 ± 6) or (0,6) and (0,-6)

and asymptotes at y= (±3/5)x 0r y= 3/5 x and y=-3/5 x

The vertices are of vertical hyperbola. The equation used will be:

\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\\

The Center of hyperbola (h,k) =(0,0)

The Distance from vertices to center is a and a = 6 (given)

For equation we have value of h,k and a and need to find value of b

we know,

y= k ± a/b (x-h)

Values of h and k are zero

y= 0 ± a/b (x-0)

y= (± a/b ) x

We are given asymtotes at y= (± 3/5)x which is equal to y= (± a/b ) x

as a = 6 then b= 10 i.e The simplified form of 6/10 is 3/5 so value of b=10

Putting values of a,b,h and k in equation we get,

\frac{(y-0)^2}{(6)^2}-\frac{(x-0)^2}{(10)^2}=1\\\\\frac{(y)^2}{36}-\frac{(x)^2}{100}=1

7 0
4 years ago
What is 5/8 as a decimal and a percent?
Anna [14]
To find the decimal divide the numerator by the denominator so
5/8=0.625 therefore as a decimal 5/8 =0.625
To find its percentage divide the numerator by the denominator and multiply by 100 so
5/8 *100=62.5 therefore as a percentage 5/8 is 62.5%

6 0
3 years ago
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