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3 years ago

What is the greatest common factor of 90,54 and 130?

Mathematics

1 answer:

Vaselesa [24]3 years ago

3 0

The answer is two. haha

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I really don't understand number 10 please help

djyliett [7]

Its asking what possible coordinates

8 0

3 years ago

Read 2 more answers

Need help, please. :)

stepladder [879]

Answer:

$x \leqslant - 3$

Step-by-step explanation:

i don't have time to do the line srry hope this helped

4 0

2 years ago

Suppose that a customer service center claims that the telephone calls it receives last 112 seconds on average, with a standard

adoni [48]

Answer:

95% Confidence interval: (111.02,112.98)

Step-by-step explanation:

Average = 112 seconds

Standard Deviation = 9 seconds

n = 324

Confidence interval:

$\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.96$

$112 \pm 1.96(\frac{9}{\sqrt{324}} ) = 112 \pm 0.98 = (111.02,112.98)$

5 0

3 years ago

Find the difference: 25 kg 32 g - 23 kg 83 g

TEA [102]

Answer:

-kg^33 *(23g^51 - 25)

Step-by-step explanation:

5 0

3 years ago

Find the standard form of the equation of the hyperbola satisfying the given conditions: X intercept +/- 6; foci at (-10,0) and

uysha [10]

Answer:

$\frac{x^{2}}{36} - \frac{y^{2}}{64}=1$

Step-by-step explanation:

Given an hyperbola with the following conditions:

Foci at (-10,0) and (10,0)
x-intercept +/- 6;

The following holds:

The center is midway between the foci, so the center must be at (h, k) = (0, 0).

The foci are 10 units to either side of the center, so c = 10 and $c^2 = 100$
The center lies on the origin, so the two x-intercepts must then also be the hyperbola's vertices.

Since the intercepts are 6 units to either side of the center, then a = 6 and $a^2 = 36.$

$Then, a^2+b^2=c^2\\b^2=100-36=64$

Therefore, substituting $a^2 = 36.$ and $b^2=64$ into the standard form

$\frac{x^{2}}{a^2} - \frac{y^{2}}{b^2}=1\\We \: have:\\ \dfrac{x^{2}}{36} - \dfrac{y^{2}}{64}=1$

4 0

3 years ago