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ZanzabumX [31]
3 years ago
11

If f(x)=11X+7, then f^-1(x)=

Mathematics
1 answer:
EleoNora [17]3 years ago
3 0

Answer:

Step-by-step explanation:

Hello!

f^{-1}(x) =\frac{1}{11x+7}

Good luck!

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I think I’m just dumb
Helga [31]

Answer: 6.2

Step-by-step explanation: you can just take 8.4 and subtract 2.2 because 2.2 is negative

8 0
3 years ago
Can someone explain this and give the answer please asap.
Artyom0805 [142]

Answer:

x=20

Step-by-step explanation:

we know the straight angle theorum...

3x+80+2x=180

5x=100

x=20

3 0
3 years ago
Read 2 more answers
Find all real numbers t such that (2/3)t - 1 < t + 7 ≤ -2t + 15. Give your answer as an interval.
Andrews [41]
Break up chain with AND statement:

(2/3)t - 1 < t + 7 AND <span>t + 7 ≤ -2t + 15

left side, we solve for t
</span>
<span>(2/3)t - 1 < t + 7
(2/3)t - t < 7 + 1
(-1/3)t < 8
t > -24

right side, we solve for t</span>
<span>t + 7 ≤ -2t + 15
t + 2t </span><span>≤ 15 - 7
3t </span><span>≤ 8
t </span><span>≤ 8/3

So our answer would be </span>
t > -24 AND t <span>≤ 8/3

as an interval, this is
</span>-24 < t <span>≤ 8/3
or in interval notation
(-24, 8/3]</span>
8 0
4 years ago
Super Bowl XLVI was played between the New York Giants and the New England Patriots in Indianapolis. Due to a decade-long rivalr
Diano4ka-milaya [45]

Answer:

Probability that from a sample of 200 Indianapolis residents, fewer than 170 were rooting for the Giants in Super Bowl XLVI is 0.02385.

Step-by-step explanation:

We are given that Due to a decade-long rivalry between the Patriots and the city's own team, the Colts, most Indianapolis residents were rooting heartily for the Giants. Suppose that 90% of Indianapolis residents wanted the Giants to beat the Patriots.

Let p = % of Indianapolis residents wanted the Giants to beat the Patriots = 90%

The z-score probability distribution for proportion is given by;

                   Z = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where,  \hat p = % of Indianapolis residents who were rooting for the Giants in Super Bowl XLVI in a sample of 200 residents = \frac{170}{200} = 0.85

           n = sample of residents = 200

So, probability that from a sample of 200 Indianapolis residents, fewer than 170 were rooting for the Giants in Super Bowl XLVI is given by = P(\hat p < 0.85)

     P(\hat p < 0.85) = P( \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < \frac{0.85-0.90}{\sqrt{\frac{0.85(1-0.85)}{200} } } ) = P(Z < -1.98) = 1 - P(Z \leq 1.98)

                                                                   = 1 - 0.97615 = 0.02385

<em>The above probability is calculated using z table by looking at value of x = 01.98 in the z table which have an area of 0.97615.</em>

<em />

Therefore, probability that fewer than 170 were rooting for the Giants in Super Bowl XLVI is 0.02385.

7 0
3 years ago
What is x*24-34+56? And how did you get the answer?
Ilia_Sergeevich [38]
I assume you are solving for x

<span>x*24-34+56 = 0

Start by isolating the term with x in it on one side of the equation.

To do this, we can subtract 56 and add 34 to both sides.

24x = -22

Divide both sides by 24 to complete the problem.

x= -22/24</span>
5 0
4 years ago
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