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2 years ago

8

You roll a fair six-sided die twice. Find the probability of rolling a 2 the first time and a number greater than 6 the second t

ime.
can you explain the steps

1 answer:

borishaifa [10]2 years ago

8 0

Answer:

the number 2 is present on one side of the six-sided die giving you a 1 in 6 chance of rolling a 2.

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Find x so that L is parallel to m

kirill [66]

Answer: y=16

Step-by-step explanation:

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3 years ago

Aiden estimates the length is 7.2 but it’s 6.5 what’s the percent error

IrinaVladis [17]

Percent error is 10.7%

Step-by-step explanation:

We need to find the percent error:

We are given:

Estimate value = 7.2

Actual value = 6.5

Percent Error=?

The formula used is:

$Percent\,\,Error=\frac{Estimate\,\,value-Actual\,\,value}{Actual\,\,value}\times100\\$

Putting values:

$Percent\,\,Error=\frac{7.2-6.5}{6.5}\times100\\Percent\,\,Error=\frac{0.7}{6.5}\times100\\Percent\,\,Error=0.107\times100\\Percent\,\,Error=10.7%\\$

So, Percent error is 10.7%

Keywords: Percent Error

Learn more about Percent Error at:

#learnwithBrainly

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3 years ago

What are the excluded values of the function? y=2/(x^2-4)

Tresset [83]

Excluded values are any values of x that make the denominator equal zero. In this case, -2 and 2 are excluded values.

3 0

3 years ago

Assume that y varies inversely with x. If y=1/3 when x=1/2, find y when x=1/4.

LenKa [72]

Answer:

y = $\frac{2}{3}$

Step-by-step explanation:

Given that y varies inversely with x then the equation relating them is

y = $\frac{k}{x}$ ← k is the constant of variation

To find k use the condition y = $\frac{1}{3}$ when x = $\frac{1}{2}$ , thus

$\frac{1}{3}$ = $\frac{k}{\frac{1}{2} }$ = 2k ( divide both sides by 2 )

k = $\frac{1}{6}$

y = $\frac{1}{6x}$ ← equation of variation

When x = $\frac{1}{4}$ , then

y = $\frac{1}{6(\frac{1}{4}) }$ = $\frac{1}{\frac{3}{2} }$ = $\frac{2}{3}$

3 0

3 years ago

Help. I need help with these questions ( see image).<br> Please show workings.

Andrew [12]

9514 1404 393

Answer:

4) 6x

5) 2x +3

Step-by-step explanation:

We can work both these problems at once by finding an applicable rule.

$\text{For $f(x)=ax^n$}\\\\\lim\limits_{h\to 0}\dfrac{f(x+h)-f(x)}{h}=\lim\limits_{h\to 0}\dfrac{a(x+h)^n-ax^n}{h}\\\\=\lim\limits_{h\to 0}\dfrac{ax^n+anx^{n-1}h+O(h^2)-ax^n}{h}=\boxed{anx^{n-1}}$

where O(h²) is the series of terms involving h² and higher powers. When divided by h, each term has h as a multiplier, so the series sums to zero when h approaches zero. Of course, if n < 2, there are no O(h²) terms in the expansion, so that can be ignored.

This can be referred to as the <em>power rule</em>.

Note that for the quadratic f(x) = ax^2 +bx +c, the limit of the sum is the sum of the limits, so this applies to the terms individually:

lim[h→0](f(x+h)-f(x))/h = 2ax +b

__

4. The gradient of 3x^2 is 3(2)x^(2-1) = 6x.

5. The gradient of x^2 +3x +1 is 2x +3.

__

If you need to "show work" for these problems individually, use the appropriate values for 'a' and 'n' in the above derivation of the power rule.

3 0

3 years ago