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2 years ago

Which ratio is not equivalent to the other three? A 2 3 B 12 15 C 18 27 D 6 9

Mathematics

1 answer:

finlep [7]2 years ago

4 0

Answer: B= 12 15

Step-by-step explanation:

The given ratios are

A) 2 3= 2: 3 already in its simplest form

B) 12 15 =12:15 Changing to its simplest form, we divide both by 3 giving us 4:5

C) 18 27 =18: 27 Changing to its simplest form, we divide both numbers by 9 giving us =2:3

D)6 9=6:9 Changing to its simplest form, we divide both by 3 giving us

2:3

Therefore the ratio not equivalent to other three is 12:15

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Seven more than the product of a number and <br> 8<br> is equal to <br> 5

yuradex [85]

8x + 7 = 5

A "product" is the result of a multiplication operation. So the product of a number, ( which we will call x,) and 8 would be 8x. (Or 8*"x" number) 7 more than that product would be 8x + 7 = 5

Hope that this helps!

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2 years ago

The mayor of a town has proposed a plan for the construction of a new community. A political study took a sample of 800 voters i

Hunter-Best [27]

Answer:

A political strategist wants to test the claim that the percentage of residents who favor construction is more than 30%, so then that represent our claim and needs to be on the alternative hypothesis.

Based on this the correct system of hypothesis are:

Null hypothesis: $p \leq 0.3$

Alternative hypothesis $p >0.3$

Step-by-step explanation:

We have the following info given from the problem:

$n= 800$ the random sample of voters selected from the town

$\hat p = 0.34$ represent the proportion of residents favored construction

$p_o = 0.30$ represent the value desired to test.

Based on this the correct system of hypothesis are:

Null hypothesis: $p \leq 0.3$

Alternative hypothesis $p >0.3$

And in order to test this hypothesis we can use a one sample z test for a population proportion and the statistic would be given by:

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

And with the data given we have:

$z=\frac{0.34 -0.3}{\sqrt{\frac{0.3(1-0.3)}{800}}}=2.469$

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2 years ago

Ex 2.8<br> 3. find the maximum value of y for the curve y=x^5 -3 for -2≤x≤1

harkovskaia [24]

$y=x^5-3\\ y'=5x^4\\\\ 5x^4=0\\ x=0\\ 0\in [-2,1]\\\\ y''=20x^3\\\\
y''(0)=20\cdot0^3=0$

The value of the second derivative for $x=0$ is neither positive nor negative, so you can't tell whether this point is a minimum or a maximum. You need to check the values of the first derivative around the point.
But the value of $5x^4$ is always positive for $x\in\mathbb{R}\setminus \{0\}$ . That means at $x=0$ there's neither minimum nor maximum.
The maximum must be then at either of the endpoints of the interval $[-2,1]$ .
The function $y$ is increasing in its entire domain, so the maximum value is at the right endpoint of the interval.

$y_{max}=y(1)=1^5-3=-2$

4 0

2 years ago