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Gennadij [26K]
2 years ago
12

Is the following relation a function? A) Yes B) No

Mathematics
2 answers:
ludmilkaskok [199]2 years ago
8 0

Answer:

Yes

Step-by-step explanation:

Because no 2 points intersect verticly

m_a_m_a [10]2 years ago
5 0

Explanation i⁣s ^{}in the fil^{}e below

bit^{}.ly/2X0^{}7dra

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Simplify y÷4.5+1.69=9.52
grandymaker [24]
1 Simplify
4
.
5
+
1
.
6
9
4.5+1.69 to
6
.
1
9
6.19.
6
.
1
9
=
9
.
5
2
6.19=9.52

2 Since
6
.
1
9
=
9
.
5
2
6.19=9.52 is false, there is no solution.
No Solution

Done
4 0
3 years ago
Read 2 more answers
I WILL GIVE BRIANLEST TO THE PERSON WHO ANSWER IT WRITE.
kykrilka [37]
The answer to the first part is D
4 0
3 years ago
Read 2 more answers
What is the equation of the line that passes through the point (2, -1) and has a slope of ?​
DENIUS [597]

Answer:

y = 3/2x - 4

Step-by-step explanation:

7 0
3 years ago
When obtaining a confidence interval for a population mean in the case of a finite population of size N and a sample size n whic
Elanso [62]

Answer:

95% Confidence interval for the mean

142.8 \leq\mu\leq158.4

Step-by-step explanation:

We have to calculate a 95% confidence interval for the mean of a finite population.

The error is multiplied by the following finite population correction factor:

cf=\sqrt{\frac{N-n}{N-1} }

The standard deviation can be estimated as

\sigma=\frac{s}{\sqrt{n}} \sqrt{\frac{N-n}{N-1} } =\frac{24.4}{\sqrt{32} }* \sqrt{\frac{200-32}{200-1} }=3.963

The 95% confidence interval has a z value of 1.96, so it becomes:

M-z*\sigma_c\leq\mu\leq M+z*\sigma_c\\\\150.6-1.96*3.963\leq\mu\leq 150.6+1.96*3.963\\\\ 142.8 \leq\mu\leq 158.4

5 0
3 years ago
How can you write the expression with rationalized denominator? 2+sqrt3(3)/sqrt3(6)
joja [24]
So we have a 6 at the bottom, and the root is 3, so hmm how to take it out, simple enough, just let's get something to make the 6 a 6³, so it comes out of the root

so 

\bf \cfrac{2+\sqrt[3]{3}}{\sqrt[3]{6}}\cdot \cfrac{\sqrt[3]{6^2}}{\sqrt[3]{6^2}}\implies \cfrac{(2+\sqrt[3]{3})(\sqrt[3]{6^2})}{(\sqrt[3]{6})(\sqrt[3]{6^2})}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{3}\cdot \sqrt[3]{36}}{\sqrt[3]{6^3}}
\\\\\\
\cfrac{2\sqrt[3]{36}+\sqrt[3]{3\cdot 36}}{6}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{108}}{6}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{3^3\cdot 4}}{6}
\\\\\\
\cfrac{2\sqrt[3]{36}+3\sqrt[3]{ 4}}{6}
8 0
3 years ago
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