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salantis [7]
3 years ago
11

Is this a function Find the domain and range

Mathematics
1 answer:
Novay_Z [31]3 years ago
8 0

Answer:

Yes

Domain: 1, -1, 0, -3, 3

Range: 1, -1, 6

Step-by-step explanation:

The domain is in the left oval, the range is in the right oval. To find if its a function you have to make sure the domain (left side) doesn't "cheat". One number in the domain cant have more than one number in the range (right side), BUT (for future reference) the range can cheat. As shown, the right side numbers 1 and 6 have two different arrow pointed at them. As long as the domain doesnt "cheat" it is a function

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Can anyone help me with this equation? I’m so lost. Thanks in advance
jeyben [28]

The measure of angle  <em>x </em> is 31 degrees.

See the attached drawing.

Start at the bottom, at the given 59° angle, and work your way up in the direction toward the angle  <em>x </em>.

6 0
3 years ago
Subtract. 5/1/10-2 9/10 Write your answer as a mixed number in simplest form.
Ahat [919]

Answer:

= 2 \frac{1}{5}

Step-by-step explanation:

5 \frac{1}{10}  - 2 \frac{9}{10}  \\  \frac{51}{10}  -  \frac{29}{10}  \\  =  \frac{22}{10}  \\  = 2 \frac{2}{10}  \\  = 2 \frac{1}{5}

hope this helps

brainliest appreciated

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8 0
3 years ago
Due tomorrow! Please
svp [43]
The answer is 79 because 9/10 divided by 71.1 is 79. And if you want to check, you can multiply 9/10 by 79, which will give you 71.1. I hope this helped :)
8 0
3 years ago
Simplify the following expression (3x^2y)^3
kolezko [41]

Answer:

\large\boxed{\left(3x^2y\right)^3=27x^6y^3}

Step-by-step explanation:

\text{Use}\ (ab)^n=a^nb^n\ \text{and}\ (a^n)^m=a^{nm}\\\\\left(3x^2y\right)^3=3^3(x^2)^3y^3=27x^6y^3

6 0
3 years ago
According to a Pew Research Center, in May 2011, 35% of all American adults had a smart phone (one which the user can use to rea
Veronika [31]

Answer:

p_v =P(z>1.82)=0.034  

Assuming a standard significance level of \alpha=0.05 the best conclusion for this case would be:

4. There is not enough evidence to show that more than 35% of community college students own a smart phone (P-value = 0.034).

Because p_v

If we select a significance level lower than 0.034 then the conclusion would change.

Step-by-step explanation:

Data given

n=300 represent the random sample taken

X=120 represent the people who have a smart phone

\hat p=\frac{120}{300}=0.4 estimated proportion of people who have a smart phone

p_o=0.35 is the value that we want to test

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

System of hypothesis

We need to conduct a hypothesis in order to test the claim that the true proportion of people who have a smart phone is higher than 0.35, the system of hypothesis are.:  

Null hypothesis:p\leq 0.35  

Alternative hypothesis:p > 0.35  

The statistic is given by:

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.4 -0.35}{\sqrt{\frac{0.35(1-0.35)}{300}}}=1.82  

Statistical decision  

Since is a right tailed test the p value would be:  

p_v =P(z>1.82)=0.034  

Assuming a standard significance level of \alpha=0.05 the best conclusion for this case would be:

4. There is not enough evidence to show that more than 35% of community college students own a smart phone (P-value = 0.034).

Because p_v

If we select a significance level lower than 0.034 then the conclusion would change.

4 0
3 years ago
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