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4 years ago

I need someone help please

Mathematics

2 answers:

DedPeter [7]4 years ago

6 0

Hi, how are you.
Question 2 is C
Question 3 is B
Hope this helps :)

Arte-miy333 [17]4 years ago

3 0

Question 2: electron and proton
and how is this math?

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Without evaluating the expression, explain why your answer is positive or negative.

Ostrovityanka [42]

Answer:

eke94 p ri494k4j4ji4i4i

idjrjrjrjrjr4j

Step-by-step explanation:

kekejdidjxxmsjs

3 0

4 years ago

A random sample of 384 people in a mid-sized city (city one) revealed 112 individuals who worked at more than one job. A second

Svetlanka [38]

Answer:

99% confidence interval is:

(0.00278 < P1 - P2< 0.15921)

Step-by-step explanation:

For calculating a confidence intervale for the difference between the proportions of workers in the two cities, we calculate the following:

$[(p_{1} - p_{2}) \pm z_{\alpha/2} \sqrt{\frac{p_{1}(1-p_{1})}{n_{1}} + \frac{p_{2}(1-p_{2})}{n_{2}} }$

Where $p_{1}$ : proportion sample of individuals who worked

at more than one job in the city one

$n_{1}$ : Number of respondents in the city one

$p_{1}$ : proportion sample of individuals who worked

at more than one job in the city two

$n_{1}$ : Number of respondents in the city two

Then

α = 0.01 and α/2 = 0.005

and $z_{\alpha/2} = 2.575$

$p_{1} = \frac{112}{384} = 0.2916$

$p_{2} = \frac{91}{432} = 0.2106$

$n_{1}= 384$ and $n_{2}= 432$

The confidence interval is:

$[(0.2916 - 0.2106) \pm 2.575 \sqrt{\frac{0.2916(1-0.2916)}{384} + \frac{0.2106(1-0.2106)}{432} }$

(0.00278 < P1 - P2< 0.15921)

5 0

3 years ago

There were 10 baked goods in a basket. four of them were sold. write a fraction to show the part of the baked goods that were no

Zina [86]

The fraction of baked goods not sold is 6/10. The numerator is the number of left over baked goods (10-4 -6) and the denominator is the total loaves of bread (10). An equivalent fraction is 12/20. This was obtained by multiplying the original fraction by two (6 x2 = 12 and 10 x2 =20).

5 0

3 years ago

Graph the system of inequalities presented here on your own paper, then use your graph to answer the following questions: y <

levacccp [35]

Solution-

The given inequalities are,

$y < 2x-7,\\\\y\geq-\frac{1}{2}x+3$

Part A

As the former inequality is only '<' so it will be represented as dotted line, but in case of the later inequality there is '≥' so it will be represented as a solid line.

This happens because in only '<, >' the values must be either lesser or greater than y, they can't take the values of y, but in case of '≤, ≥' the values can be equal to y.

Taking origin (0, 0) and putting in the first inequality.

$\Rightarrow 0< 2(0)-7\\\\\Rightarrow 0\nless -7$

As the equation does not satisfy, so the shaded area will be on the other side of the origin.

Again taking origin (0, 0) and putting in the second inequality.

$\Rightarrow 0\geq-\frac{1}{2}0+3\\\\\Rightarrow 0\ngeqslant 3$

As the equation does not satisfy, so the shaded area will be on the other side of the origin.

The second attachment contains the feasible region.

Part B

In order to know whether the point (3, −7) is included in the solution area for the system or not, we must put it in both the equations, if it satisfies both then only it is within the solution area.

$\Rightarrow -7< 2(3)-7,\ -7\geq-\frac{1}{2}(3)+3\\\\\Rightarrow -7$

As the point (3, -7) only satisfies first inequality, but not the second inequality so this point won't be in the solution area.

6 0

3 years ago

Read 2 more answers

Divide (x^2+9x+19)÷(x+5)

gayaneshka [121]

Since this can’t be factored easily, you have to use synthetic division. This makes the answer x+4+(-1/x+5)

6 0

4 years ago