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3 years ago

Convert the following using dimensional analysis: 13.2 cm to ft & in

1 answer:

6 0

Multiply (13.2 cm) by (1 inch/2.54 cm). That gives you the whole length in inches. If it's more than 12, then take as many 12s away from it as you can, and call each of those '1 foot'. When you can't take 12 away any more, then what you have left is the inches.

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Enter the value of b when the expression 14.1x - b is equivalent 4.7(3x-3.5)?

3241004551 [841]

Answer:

I believe it is: B= 16.45

4 0

3 years ago

One canned juice is 15% orange juice another 10% orange juice. How many liters of each should be mixed together in order to get

Delicious77 [7]

<span>Equation:
0.15x + 0.10(5-x) = 0.11*5
---
Multiply thru by 100:
15x + 10*5 - 10x = 11*5

---
5x = 5
x = 1 liter (amt. 15% oj needed)
----
5-x = 4 liters (amt of 10% oj needed)
======</span>

3 0

3 years ago

write the standard equation of a circle that is tangent to the x-axis with the center located at (2,4)

dmitriy555 [2]

Answer:

Step-by-step explanation:

radius=4 as it is tangent to x-axis.

y co-ordinate of center is 4 so radius=4

eq. of circle is (x-2)²+(y-4)²=4²

or x²-4x+4+y²-8y+16=16

or x²+y²-4x-8y+4=0

4 0

3 years ago

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

Suppose the mean SAT verbal score is 525 with a standard deviation of 100, while the mean SAT math score is 575 with a standard

Ipatiy [6.2K]

Answer:

The mean of the combined math and verbal scores is 1100, while the standard deviation is 141.

Step-by-step explanation:

Normal variables

Normal variables have mean $\mu$ and standard deviation $\sigma$

When we add normal variables, the combined mean is the sum of both means, and the standard deviation is the square root of the sum of both variances. The distribution is still normal.

In this question:

Verbal: $\mu_{V} = 525, \sigma_{V} = 100$ .

Math: $\mu_{M} = 575, \sigma_{M} = 100$

Combined:

$\mu = \mu_{V} + \mu_{M} = 525 + 575 = 1100$

$\sigma = \sqrt{\sigma_{V}^{2}+\sigma_{M}^{2}} = \sqrt{100^2 + 100^2} = 141$

The mean of the combined math and verbal scores is 1100, while the standard deviation is 141.

6 0

3 years ago