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2 years ago

Please help I'm in desperate need of assistance

Mathematics

1 answer:

ivanzaharov [21]2 years ago

7 0

Answer:

Step-by-step explanation: I think for figure 0 is 7.figure 1 is 9 and the growth is for every increase of the figure number it adds 2 for the tile number.

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The scores of an eighth-grade math test have a normal distribution with a mean Mu = 83 and a standard deviation Sigma = 5. If Di

tresset_1 [31]

The missing choices are:

A. z-score = 92 - 83

B. z-score = (92 - 83)/5

C. z-score = 83 - 92

D. z-score = (92 - 83)/92

The expression she would write to find z-score is z-score = $\frac{92-83}{5}$ ⇒ B

Step-by-step explanation:

The rule of z-score is z = (x -μ)/σ, where

x is the score
μ is the mean
σ is the standard deviation

The scores of an eighth-grade math test have a normal distribution

∵ The mean is 83

∴ μ = 83

∵ The standard deviation is 5

∴ σ = 5

∵ Din’s test score was 92

∴ x = 92

- Substitute all of these values in the formula of z-score

∴ z-score = $\frac{92-83}{5}$

The expression she would write to find z-score is z-score = $\frac{92-83}{5}$

Learn more:

You can learn more about z-score in brainly.com/question/7207785

#LearnwithBrainly

5 0

2 years ago

Pls Help plsssssssss

Korvikt [17]

One solution would be the answer

7 0

3 years ago

Match each value with its formula for ABC.

MariettaO [177]

The solution to the question is:

c is 6 = $\sqrt{a^{2} + b^{2} -2abcosC }$

b is 5 = $\sqrt{a^{2} + c^{2} -2accosB }$

cosB is 2 = $\frac{a^{2} + c^{2} - b^{2} }{2ac}$

a is 4 = $\sqrt{b^{2} + c^{2} -2bccosA }$

cosA is 3 = $\frac{b^{2} + c^{2} -a^{2} }{2bc}$

cosC is 1 = $\frac{b^{2} + a^{2} - c^{2} }{2ab}$

<h3>What is cosine rule?</h3>

it is used to relate the three sides of a triangle with the angle facing one of its sides.

The square of the side facing the included angle is equal to the some of the squares of the other sides and the product of twice the other two sides and the cosine of the included angle.

Analysis:

If c is the side facing the included angle C, then

$c^{2}$ = $a^{2}$ + $b^{2}$ -2ab cos C-----------------1

then c = $\sqrt{a^{2} + b^{2} -2abcosC }$

if b is the side facing the included angle B, then

$b^{2}$ = $a^{2}$ + $c^{2}$ -2accosB-----------------2

b = $\sqrt{a^{2} + c^{2} -2accosB }$

from equation 2, make cosB the subject of equation

2ac cosB = $a^{2}$ + $c^{2}$ - $b^{2}$

cosB = $\frac{a^{2} + c^{2} - b^{2} }{2ac}$

if a is the side facing the included angle A, then

$a^{2}$ = $b^{2}$ + $c^{2}$ -2bccosA--------------------3

a = $\sqrt{b^{2} + c^{2} -2bccosA }$

from equation 3, making cosA subject of the equation

2bcosA = $b^{2}$ + $c^{2}$ - $a^{2}$

cosA = $\frac{b^{2} + c^{2} -a^{2} }{2bc}$

from equation 1, making cos C the subject

2abcosC = $b^{2}$ + $a^{2}$ - $c^{2}$

cos C = $\frac{b^{2} + a^{2} - c^{2} }{2ab}$

In conclusion,

c is 6 = $\sqrt{a^{2} + b^{2} -2abcosC }$

b is 5 = $\sqrt{a^{2} + c^{2} -2accosB }$

cosB is 2 = $\frac{a^{2} + c^{2} - b^{2} }{2ac}$

a is 4 = $\sqrt{b^{2} + c^{2} -2bccosA }$

cosA is 3 = $\frac{b^{2} + c^{2} -a^{2} }{2bc}$

cosC is 1 = $\frac{b^{2} + a^{2} - c^{2} }{2ab}$

Learn more about cosine rule: brainly.com/question/4372174

$SPJ1

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2 years ago

Solve the inequality

Mamont248 [21]

32 and Konner is right please

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2 years ago

If I get a 90 on a quiz and I have a 88.4 in the class will it bring the grade up

g100num [7]

It should but 88 is still a A

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3 years ago