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3 years ago

Please help me find the values!!! I need to submit this soon

Mathematics

1 answer:

Gre4nikov [31]3 years ago

3 0

Answer:

b = 2

c = 2√2

Step-by-step explanation:

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Which angles are corresponding angles?

Reptile [31]

Answer:

1 and 3

15 and 13

15 and 7

1 and 9

Step-by-step explanation:

5 0

3 years ago

Parallelogram PQRS is shown. m/S=(3x), m/R=120, and m/Q=(2y). Find the value of y.A.

fomenos

Option B. From the parallelogram PQRS the value of y is given to be 30

<h3>How to solve for the value of y from the parallelogram</h3>

In order to get the value of y we have to use the formula

2y + 120 = 80

where the value 120 is the angle that is stated as 120 from the question

2y = 180 - 120

2y = 60

y = 60 / 2

y = 30

Hence the value of y = 30

We can go ahead to get the value of x as well

3x + 120 = 180

take the like terms

3x = 180 - 120

3x = 60

divide through by 3 to get x

60 / 3 = x

20 = x

Read more on parallelograms here: brainly.com/question/24056495

#SPJ1

7 0

1 year ago

A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 student

yan [13]

Answer:

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

Step-by-step explanation:

The (1 - α)% confidence interval for the difference between population means is:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

The information provided is as follows:

$n_{1}= 138\\n_{2}=156\\\bar x_{1}=61\\\bar x_{2}=64.6\\\sigma_{1}=18.53\\\sigma_{2}=13.43$

The critical value of z for 98% confidence level is,

$z_{\alpha/2}=z_{0.02/2}=2.326$

Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

$=(61-64.6)\pm 2.326\times\sqrt{\frac{(18.53)^{2}}{138}+\frac{(13.43)^{2}}{156}}\\\\=-3.6\pm 4.4404\\\\=(-8.0404, 0.8404)\\\\\approx (-8.04, 0.84)$

Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

5 0

3 years ago

Rewrite the slope intercept equation of the line y=1/3x-2 in standard form

kaheart [24]

y=1/3 x - 2
3y = x - 6
x - 3y = 6 ...this is standard form

7 0

3 years ago

Do not count the individual oranges. count them as groups.

kupik [55]

Answer:

A. 5

B. 100

C. 300

4 0

3 years ago