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hichkok12 [17]
2 years ago
9

The graph of a function is shown below. find the following, g(10), g(-3)

Mathematics
1 answer:
Rzqust [24]2 years ago
4 0

Answer:

g(5x)=0

Step-by-step explanation:

Given that g(x)=4

From the graph when x=-2 then g(x)=4

Hence 5x=-10

Now from the graph

g(-10)=0

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If there was 180 keys and he sent some and now he got 107 keys how many did he spend
Black_prince [1.1K]

Answer:

73

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Two sets of equatic expressions are shown below in various forms: Line 1: x2 + 3x + 2 (x + 1)(x + 2) (x + 1.5)2 − 0.25 Line 2: x
kherson [118]

Answer:  The correct line is

\textup{Line 1 :}x^2+3x+2=(x+1)(x+2)=(x+1.5)^2-0.25.

Step-by-step explanation:  We are given the following two sets of quadratic expressions in various forms:

\textup{Line 1: }x^2+3x+2=(x+1)(x+2)=(x+1.5)^2-0.25,\\\\\textup{Line 2 :}x^2+5x+6=(x+2)(x+3)=(x+2.5)^2+6.25.

We are to select one of the lines from above that represent three equivalent expressions.

We can see that there are three different forms of a quadratic expression in each of the lines:

First one is the simplified form, second is the factorised form and third one is the vertex form.

So, to check which line is correct, we need to calculate the factorised form and the vertex form from the simplified form.

We have

\textup{Line 1: }\\\\x^2+3x+2\\\\=x^2+2x+x+2\\\\=x(x+2)+1(x+2)\\\\=(x+1)(x+2),

and

x^2+3x+2\\\\=x^2+2\times x\times 1.5+(1.5)^2-(1.5)^2+2\\\\=(x+1.5)^2-2.25+2\\\\=(x+1.5)^2-0.25.

So,

\textup{Line 1 :}x^2+3x+2=(x+1)(x+2)=(x+1.5)^2-0.25.

Thus, Line 1 contains three equivalent expressions.

Now,

\textup{Line 2: }\\\\x^2+5x+6\\\\=x^2+3x+2x+6\\\\=x(x+3)+2(x+3)\\\\=(x+2)(x+3),

and

x^2+5x+6\\\\=x^2+2\times x\times 2.5+(2.5)^2-(2.5)^2+6\\\\=(x+2.5)^2-6.25+6\\\\=(x+2.5)^2-0.25\neq (x+2.5)^2+6.25.

So,

\textup{Line 2 :}x^2+3x+2=(x+1)(x+2)=(x+1.5)^2+6.25.

Thus, Line 2 does not contain three equivalent expressions.

Hence, Line 1 is correct.

7 0
3 years ago
Two random samples are taken, with each group asked if they support a particular candidate. A summary of the sample sizes and pr
Minchanka [31]

Answer:

And we got \alpha/2 =0.01 so then the value for \alpha=0.02 and then the confidence level is given by: Conf=1-0.02=0.98[/tex[ or 98%Step-by-step explanation:A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  The margin of error is the range of values below and above the sample statistic in a confidence interval.  Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  [tex]p_1 represent the real population proportion for 1

\hat p_1 =0.768 represent the estimated proportion for 1

n_1=92 is the sample size required for 1

p_2 represent the real population proportion for 2

\hat p_2 =0.646 represent the estimated proportion for 2

n_2=95 is the sample size required for 2

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

The confidence interval for the difference of two proportions would be given by this formula  

(\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}  

For this case we have the confidence interval given by: (-0.0313,0.2753). From this we can find the margin of erro on this way:

ME= \frac{0.2753-(-0.0313)}{2}=0.1533

And we know that the margin of erro is given by:

ME=z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}

We have all the values except the value for z_{\alpha/2}

So we can find it like this:

0.1533=z_{\alpha/2} \sqrt{\frac{0.768(1-0.768)}{92} +\frac{0.646 (1-0.646)}{95}}

And solving for z_{\alpha/2} we got:

z_{\alpha/2}=2.326

And we can find the value for \alpha/2 with the following excel code:

"=1-NORM.DIST(2.326,0,1,TRUE)"

And we got \alpha/2 =0.01 so then the value for \alpha=0.02 and then the confidence level is given by: Conf=1-0.02=0.98 or 98%

7 0
3 years ago
Slope is -2 and (1,1) is on the line
Mrac [35]
-2x+3. X intercept is +3. Moves two steps down and one step to the right to land on the point (1,1)
7 0
3 years ago
what is the difference between the mean and the median of the following distribution? {1,1,1,2,2,2,2,3,3,4,4,4,4,5,5,6,7,7,8,8,9
iren2701 [21]

The given distribution =

1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 5, 5, 6, 7, 7, 8, 8, 9

To find mean, we have to add the given numbers and divide it by the total number of terms given.

Here 21 total numbers given.

Mean = (1+1+1+2+2+2+2+3+3+4+4+4+4+5+5+6+7+7+8+8+9) /21

= (88)/21 = 4.19 (Approximate upto two decimal place)

Median is the middle value of the distribution.

1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 5, 5, 6, 7, 7, 8, 8, 9

There are 21 numbers given. So middle value is the number in the 11th position.

The number in the 11th position is 4.

So, the median of the distribution = 4

We have to find the difference between mean and median here. To find the difference we have to subtract mean and median.

Therefore, the required difference = 4.19 - 4 = 0.19

0.19 is the required answer here.

So, option b is correct.

3 0
3 years ago
Read 2 more answers
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