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MArishka [77]
3 years ago
13

I need help with both of these questions please

Mathematics
1 answer:
tankabanditka [31]3 years ago
3 0

Answer:

\frac{22}{30} = \frac{11}{15}

and

\frac{176}{7} = 25 \frac{1}{7}

Step-by-step explanation:

First make all the fractions into improper fractions

1\frac{5}{6} = \frac{11}{6} \\\\2\frac{1}{2} = \frac{5}{2} \\\\3\frac{1}{7} = \frac{22}{7}

After doing that put them back into the problems

Problem 1)

1\frac{5}{6}  ÷ 2\frac{1}{2} , plug in the improper fractions

\frac{11}{6} ÷ \frac{5}{2}

to divide, you need to flip the second fraction and multiply

\frac{11}{6}· \frac{2}{5} = \frac{22}{30}

then reduce

\frac{22}{30} = \frac{11}{15}

Problem 2)

3\frac{1}{7}÷ \frac{1}{8}\\

Plug in improper fraction

\frac{22}{7} ÷ \frac{1}{8}\\

Then flip second fraction and multiply

\frac{22}{7} · \frac{8}{1}

Multiply

\frac{176}{7}

Reduce, or make it a mixed number

\frac{176}{7} = 25 \frac{1}{7}

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Jet001 [13]

a. f has an average value on [5, 11] of

f_{\rm ave}=\displaystyle\frac1{11-5}\int_5^{11}(x-7)^2\,\mathrm dx=\frac{(x-7)^3}{18}\bigg|_5^{11}=\frac{4^3-(-2)^3}{18}=4

b. The mean value theorem guarantees the existence of c\in(5,11) such that f(c)=f_{\rm ave}. This happens for

(c-7)^2=4\implies c-7=\pm2\implies c=9\text{ or }c=5

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3 years ago
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2 years ago
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Find the value of x<br><br>urgent!! plz help me! will give the brainliest​
Anarel [89]

Answer:

x = 2

Step-by-step explanation:

\frac{1}{2} (x - 1) - \frac{1}{6} (x+1) = 0

In an equation our aim is to find the value of what we are looking for as well as keeping the equation balanced. For example if we divided by 2  only from one side then the equation would change so it's an important rule to keep in mind when solving equations, that you need to keep both sides of the equation the same.

\frac{1}{2} (x - 1) - \frac{1}{6} (x+1) = 0

→ Expand the brackets

\frac{1}{2} x- \frac{1}{2}-\frac{1}{6}x -\frac{1}{6} =0

→ Multiply everything by 12 to make solving the equation easier

6x - 6 - 2x - 2 = 0

→ Simplify equation

4x - 8 = 0

→ Add 8 to both sides to isolate 4x

4x = 8

→ Divide by 4 on both sides to isolate x

x = 2

⇒ We can substitute x = 2 back into the equation to see if the solution is correct, if we get 0 on both sides then x = 2 is correct

\frac{1}{2} (x - 1) - \frac{1}{6} (x+1) = 0

⇒ Substitute in the values

\frac{1}{2} (2-1)-\frac{1}{6} (2+1) = 0

⇒ Simplify

\frac{1}{2} (1)-\frac{1}{6} (3) = 0

⇒ Simplify further

\frac{1}{2} -\frac{1}{2} =0

0 = 0

The solution x = 2 is correct

4 0
3 years ago
23 + 4 + x = 25 -12 porfa ayuda :(
MAXImum [283]

Answer:

x = - 14

Step-by-step explanation:

23 + 4 + x = 25 - 12

Combine like terms on both sides of the equation (by adding 23 and 4 on the left-hand side, and subtracting 12 from 25 on the right-hand side):

23 + 4 + x = 25 - 12

27 + x = 13

Next, subtract 27 from both sides to solve for x:

27 - 27 + x = 13 - 27

x = - 14

Please mark my answers as the Brainliest if you find this explanation helpful :)

4 0
3 years ago
A credit reporting agency claims that the mean credit card debt in a town is greater than $3500. A random sample of the credit c
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Answer:

Claim is false

Step-by-step explanation:

Claim : A credit reporting agency claims that the mean credit card debt in a town is greater than $3500.

H_0:\mu \leq 3500\\H_a:\mu = 3500

n = 20

Since n <30

So we will use t test

Formula : t =\frac{x-\mu}{\frac{s}{\sqrt{n}}}

s = standard deviation = 391

x = 3600

n = 20

t =\frac{3600-3500}{\frac{391}{\sqrt{20}}}

t =1.14

Degree of freedom = n-1 = 20-1 = 19

α=0.10

So, using t table

t_({\frac{\alpha}{2},d.f.}) = 1.72

t critical > t calculated

So we accept the null hypothesis

Hence we reject the claim that the mean credit card debt in a town is greater than $3500.

7 0
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