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2 years ago

Solve for “z.” Please help!

Mathematics

1 answer:

Darya [45]2 years ago

7 0

Answer: Z=5

Step-by-step explanation:

Solve the rational equation by combining expression and isolating the variable z

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Anyone know this, if you do plz help me. If the octagon is cm wide what is the area?

DENIUS [597]

Answer:

Step-by-step explanation:

Corner triangle: 1/2 (2 x 2) = 2

4 triangles = 4 x 2 = 8

Octagon + 4 triangles forms a square: 10 x 10 = 100

Octagon: 100 - 8 = 92

3 0

2 years ago

The time needed to complete a final examination in a particular college course is normallydistributed with a mean of 80 minutes

Artist 52 [7]

Answer:

a) 0.023

b) 0.286

c) 10 students will be unable to complete the exam inthe allotted time.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 80 minutes

Standard Deviation, σ = 10 minutes

We are given that the distribution of time to complete an exam is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(completing the exam in one hour or less)

P(x < 60)

$P( x < 60) = P( z < \displaystyle\frac{60 - 80}{10}) = P(z < -2)$

Calculation the value from standard normal z table, we have,

$P(x < 60) =0.023= 2.3\%$

b) P(complete the exam in more than 60 minutes but less than 75 minutes)

$P(60 \leq x \leq 75) = P(\displaystyle\frac{60 - 80}{10} \leq z \leq \displaystyle\frac{75-80}{10}) = P(-2 \leq z \leq -0.5)\\\\= P(z \leq -0.5) - P(z < -2)\\= 0.309- 0.023 = 0.286= 28.6\%$

c) P(completing the exam in more than 90 minutes)

P(x > 90)

$P( x > 90) = P( z > \displaystyle\frac{90 -80}{10}) = P(z > 1)$

$= 1 - P(z \leq 1)$

Calculation the value from standard normal z table, we have,

$P(x > 90) = 1 - 0.8413 = 0.1587 = 15.87\%$

15.87% of children of class will require more than 90 minutes to complete the test.

Number of children =

$\dfrac{15.87}{100}\times 60 = 9.52\approx 10$

Approximately, 10 students of class will require more than 90 minutes to complete the test.

4 0

3 years ago

How do you simplify 4/^3√4

OlgaM077 [116]

$\dfrac{4}{\sqrt[3]{4}}=\dfrac{4\cdot\sqrt[3]2}{\sqrt[3]{4}\cdot\sqrt[3]2}=\dfrac{4\sqrt[3]2}{\sqrt[3]{4\cdot2}}=\dfrac{4\sqrt[3]2}{\sqrt[3]8}=\dfrac{4\sqrt[3]2}{2}=2\sqrt[3]2$

$Use\ \sqrt[3]a\cdot\sqrt[3]b=\sqrt[3]{ab}$

7 0

2 years ago

How do u solve this problem 3 1/2 - 2/3

SashulF [63]

3 1/2 = 4/2 - 2/3
2×3 =6
12/6 -4/6 =8/6 =4/3 = 1 & 1/3

8 0

2 years ago

T2.T?

Setler [38]

9514 1404 393

Answer:

Step-by-step explanation:

Reversing the digits changes the value by a factor of 9 times the difference in the digit values. Here, this means the digits differ by 27/9 = 3. If one digit is double the other, the higher digit is 2×3 = 6.

The number is 63.

_____

If you like, you can work through the algebra of it. Let the original number have tens digit x and ones digit y.

original number value = 10x +y

reversed number value = 10y +x

difference of values = 27 = (10x +y) -(10y +x) = 9(x -y)

The tens digit is twice the ones digit, so we have ...

x = 2y

Substituting in the difference equation, we have ...

27 = 9(x -y)

3 = (2y -y) = y . . . . . divide by 9, substitute for x

x = 2(3) = 6

The number is 10x+y = 63.

4 0

2 years ago