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Ksenya-84 [330]
3 years ago
8

Pls tell me the answer with an explanation how you got it!

Mathematics
1 answer:
Flura [38]3 years ago
8 0

Answer:

25 degrees

Step-by-step explanation:

40 + 50 = 90

2 x 25 = 5

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Find the inequality represented by the graph someone help please
Alecsey [184]
U have a solid line...meaning there is an equal sign in the problem
u have a y int of -1 (the y int is where ur line crosses the y axis)
u have a slope of : 1
it is shaded below the line...so it is less then

so ur inequality is : y < = x - 1 <==

6 0
3 years ago
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Select the statement that describes this expression: (10 − 3) x 4 + 5.
tankabanditka [31]

Answer:

Multiply 4 by the difference of 10 and 3, then add 5

4 0
2 years ago
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Help, please!!! This is urgent!!!!!
Wewaii [24]

Answer:

Part A: one solution:

Part B: x = 3, y = 4.

1) Part A: how many solutions does the pair of equations for lines A and B have?

The solution of a system of equations in a graph is given by the intersetion of the curves that represent the equations.

In this case, there are two straight lines, which intersect in one and only one point.

Hence, the system has one solution.

2) Part B: what is the solution to the equations of lines A and B?

The solution is the pair of coordinates of the intersection point. It is (3, 4).

Therefore, the solution is x = 3, y = 4.

Step-by-step explanation:

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5 0
3 years ago
Mary weighs n kg while her sister weighs 15 kg less. How many times is Mary heavier than her sister?
kari74 [83]

Answer:

Mary is  \frac{n}{n-15}  times heavier than her sister

Step-by-step explanation:

Mary's Weight = n

Sister's Weight = n - 15

To find how many times mary is heavier, we need to multiply "that many time" [let it be x] with sister's weight to get Mary's weight.

<u>e.g. </u>If someone weighs 100 and another person 50, 50 * 2 = 100, thus, 100 kg person is 2 times heaving. We do same for this:

(n - 15) * x = n

x = n/(n - 15)

3 0
3 years ago
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The 2008 Workplace Productivity Survey, commissioned by LexisNexis and prepared by World One Research, included the question, "H
LenaWriter [7]

Answer:

A 95% confidence interval estimate of the mean number of hours a legal professional works on a typical workday is [7.44 hours, 10.56 hours].

Step-by-step explanation:

We are given that x is normally distributed with a known standard deviation of 12.6.

A sample of 250 legal professionals was surveyed, and the sample's mean response was 9 hours.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                               P.Q.  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample average mean response = 9 hours

            \sigma  = population standard deviation = 12.6

            n = sample of legal professionals = 250

            \mu = mean number of hours a legal professional works

<em>Here for constructing a 95% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.</em>

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                   of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < -1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ]

                                       = [ 9-1.96 \times {\frac{12.6}{\sqrt{250} } } , 9+1.96 \times {\frac{12.6}{\sqrt{250} } } ]

                                       = [7.44 hours, 10.56 hours]

Therefore, a 95% confidence interval estimate of the mean number of hours a legal professional works on a typical workday is [7.44 hours, 10.56 hours].

5 0
2 years ago
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