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3 years ago

7

Please answer me this question

1 answer:

Jobisdone [24]3 years ago

4 0

Answer:

x = 2

y = 3

Step-by-step explanation:

$\left \{ {{0.4x + 0.3y = 1.7} \atop {0.7x - 0.2y = 0.8}} \right.$

$\left \{ {{2(0.4x + 0.3y) = 2 * 1.7} \atop {3(0.7x - 0.2y) = 3 * 0.8}} \right.$

$\left \{ {{0.8x + 0.6y = 3.4} \atop {2.1x - 0.6y) = 2.4}} \right.$

$2.9x = 5.8$

$29x = 58$

$x = 2$

$0.3y + 0.4x = 1.7$

$0.3y + 0.4 * 2 = 1.7$

$0.3y + 0.8 = 1.7$

$0.3y = 0.9$

$3y = 9$

$y = 3$

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Answer:

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Step-by-step explanation:

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3 years ago

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If f(−1) = −3 and f prime of x equals the quotient of 4 times x squared and the quantity x cubed plus 3 , which of the following

lora16 [44]

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Please answer. In parallelogram WXYZ, diagonals WY and XZ intersect at point A. Give WA=x^2-48 and AY=x^2-6x. What is WY? Show y

netineya [11]

The diagonals of a parallelogram are congruent

The length WY is 32

<h3>How to determine the length WY</h3>

Given that:

$WA = x^2 - 48$ and

$AY = x^2 - 6x$

Then, we have:

$x^2 - 48 = x^2 - 6x$ diagonal

Subtract x^2 from both sides

$- 48 =- 6x$

Divide both sides by -6

$x = 8$

The length WY is calculated as:

$WY=2 * WA$

So, we have:

$WY=2 * [x^2 - 48]$

Substitute 8 for x

$WY=2 * [8^2 - 48]$

Simplify

$WY=32$

Hence, the length WY is 32

Read more about parallelograms at:

brainly.com/question/10062747

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3 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.

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7nadin3 [17]

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