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3 years ago

Can some one help me with this

Mathematics

2 answers:

N76 [4]3 years ago

7 0

I think this is how you solve this and the answer 0.1+30+8−12+14=26.35

cestrela7 [59]3 years ago

6 0

Plug in the numbers for the variables
0.1(30)+8-12(1/4)
3+8-3
8

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List the sides of triangle WXY in order from smallest to largest.

nataly862011 [7]

Answer:

Using the relation between angles and sides of any triangle the answer is:

Third option: WX, XY, YW

Step-by-step explanation:

<X=90° (right angle)

<W=51°

<Y=?

The sum of the interior angles of any triangle is 180°, then:

<W+<X+<Y=180°

Replacing the given values:

51°+90°+<Y=180°

141°+<Y=180°

Solving for <Y: Subtracting 141° both sides of the equation:

141°+<Y-141°=180°-141°

<Y=39°

The order of the angles from smallest to largest is:

<Y=39°, <W=51°, <X=90°

The opposite sides to these angles must be ordered in the same way:

Opposite side to <Y: WX

Opposite side to <W: XY

Opposite side to <X: YW

Then the order of the sides from smallest to largest is:

WX, XY, YW

6 0

4 years ago

The points obtained by students of a class in a test are normally distributed with a mean of 60 points and a standard deviation

Paraphin [41]

Answer:

0.13% of students have scored less than 45 points

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 60, \sigma = 5$

About what percent of students have scored less than 45 points?

This is the pvalue of Z when X = 45. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{45 - 60}{5}$

$Z = -3$

$Z = -3$ has a pvalue of 0.0013

0.13% of students have scored less than 45 points

3 0

4 years ago

Does anyone know the surface area for this problem. if so i will mark brainliest.

sveticcg [70]

From the given volume, we know that the ratio is:
$\text {Ratio of yellow eraser to white eraser = } \sqrt[3]{ \dfrac{16}{2} }$

From the given area, we know that the ratio is:
$\text {Ratio of yellow eraser to white eraser = } \sqrt{ \dfrac{52}{x} }$

Equate the two ratio and solve for x:
$\sqrt[3]{ \dfrac{16}{2} } = \sqrt{ \dfrac{52}{x} }$

Cube both sides:
$\dfrac{16}{2} = \bigg(\sqrt{ \dfrac{52}{x} }\bigg)^3$

Square both sides:
$\bigg(\dfrac{16}{2} \bigg)^2 = \bigg( \dfrac{52}{x} }\bigg)^3$

Simplify each term:
$\dfrac{16^2}{2^2} = \dfrac{52^3}{x^3}$

Cross multiply:
$256x^3 = 562432$

Divide both sides by 256:
$x^3 = 2197$

Cube root both sides:
$x = 13$

7 0

4 years ago

3. for each item, decide whether or not the given expression is defined. for each item that is defined, compute the result. (a)

Sati [7]

The results of given matrices can be obtained using matrix multiplication.

<h3>Find the results of the given matrices:</h3>

Here in the question it is given that,

A = $\left[\begin{array}{ccc}1&-1&2\\3&1&4\end{array}\right]$ , B = $\left[\begin{array}{ccc}2&-1&3\\5&1&2\\4&6&-2\end{array}\right]$ , C = $\left[\begin{array}{ccc}1\\-1\\2\end{array}\right]$ , D = $\left[\begin{array}{ccc}2&-2&3\end{array}\right]$ ,

E = $\left[\begin{array}{ccc}2-i&1+i\\-i&2+4i\end{array}\right]$ , F = $\left[\begin{array}{ccc}i&1-3i\\0&4+i\end{array}\right]$

We have to find AB, BC, CA, CD, $C^{T} A^{T}$ , F², $BD^{T}$ , $A^{T} A$ and FE.

AB = $\left[\begin{array}{ccc}1&-1&2\\3&1&4\end{array}\right]\left[\begin{array}{ccc}2&-1&3\\5&1&2\\4&6&-2\end{array}\right]$

a₁₁ = 1×2 + (-1)×5 + 2×4 = 5, a₁₂ = 1×(-1) + (-1)×1 + 2×6 = 10, a₁₃ = 1×3 + (-1)×2 + 2×(-2) = -3, a₂₁ = 3×2 + 1×5 + 4×4 = 27, a₂₂ = 3×(-1) + 1×1 + 4×6 = 22, a₂₃ = 3×3 + 1×2 + 4×(-2) = 3

AB = $\left[\begin{array}{ccc}5&10&-3\\27&22&3\end{array}\right]$

BC = $\left[\begin{array}{ccc}2&-1&3\\5&1&2\\4&6&-2\end{array}\right]$ $\left[\begin{array}{ccc}1\\-1\\2\end{array}\right]$

a₁₁ = 2×1 + (-1)×(-1) + 3×2 = 9, a₂₁ = 5×1 + 1×(-1) + 2×2 = 8, a₃₁ = 4×1 + 6×(-1) + (-2)×2 = -6

BC = $\left[\begin{array}{ccc}9\\8\\-6\end{array}\right]$

CA, CA is not defined since dimension of the matrices are 3×1 and 2×3

$A^{T}E = \left[\begin{array}{ccc}1&3\\-1&1\\2&4\end{array}\right]\left[\begin{array}{ccc}2-i&1+i\\-i&2+4i\end{array}\right]$

a₁₁ = 1×(2-i) + 3×(-i) = 2-4i, a₁₂ = 1x(1+i) + 3×(2+4i) = 7+13i, a₂₁ = -1×(2-i) + 1×(-i) = -2, a₂₂ = -1×(1+i) + 1×(2+4i) = 1+3i, a₃₁ = 2×(2-i) + 4×(-i) = 4-6i, a₃₂ = 2×(1+i) + 4×(2+4i) = 10+18i

$A^{T}E = \left[\begin{array}{ccc}2-4i&7+13i\\-2&1+3i\\4-6i&10+18i\end{array}\right]$

CD = $\left[\begin{array}{ccc}1\\-1\\2\end{array}\right]$ $\left[\begin{array}{ccc}2&-2&3\end{array}\right]$

a₁₁ = 1×2 = 2, a₁₂ = 1×(-2) = -2, a₁₃ = 1×3 = 3, a₂₁ = -1×2 = -2, a₂₂ = -1×(-2) = 2, a₂₃ = -1×3 = -3,a₃₁= 2×2 = 4, a₃₂ = 2×(-2) = -4, a₃₃ = 2×3 = 6

CD = $\left[\begin{array}{ccc}2&-2&3\\-2&2&-3\\4&-4&6\end{array}\right]$

$C^{T} A^{T} =\left[\begin{array}{ccc}1&-1&2\end{array}\right]\left[\begin{array}{ccc}1&3\\-1&1\\2&4\end{array}\right]$

a₁₁ = 1×1 + (-1)×(-1) + 2×2 = 6, a₁₂ = 1×3 + (-1)×1 + 2×4 = 10

$C^{T}A^{T}=\left[\begin{array}{ccc}6&10\end{array}\right]$

F² = $\left[\begin{array}{ccc}i&1-3i\\0&4+i\end{array}\right]\left[\begin{array}{ccc}i&1-3i\\0&4+i\end{array}\right]$

a₁₁ = i×i + (1-3i)×0 = -1,a₁₂ = i×(1-3i) + (1-3i)×(4+i) = 10-10i, a₂₁= 0×i + (4+i)×0 = 0, a₂₂ = 0×(1-3i) + (4+i)×(4+i) = 15+8i

F² = $\left[\begin{array}{ccc}-1&10-10i\\0&15+8i\end{array}\right]$

$BD^{T}=\left[\begin{array}{ccc}2&-1&3\\5&1&2\\4&6&-2\end{array}\right]\left[\begin{array}{ccc}2\\-2\\3\end{array}\right]$

a₁₁ = 2×2 + (-1)×(-2) + 3×3 = 15, a₂₁ = 5×2 + 1×(-2) + 2×3 = 14, a₃₁ = 4×2 + 6×(-2) + (-2)×3 = -10

$BD^{T}= \left[\begin{array}{ccc}15\\14\\-10\end{array}\right]$

$A^{T} A=\left[\begin{array}{ccc}1&3\\-1&1\\2&4\end{array}\right] \left[\begin{array}{ccc}1&-1&2\\3&1&4\end{array}\right]$

a₁₁ = 1×1 + 3×3 = 10, a₁₂ = 1×(-1) + 3×1 = 2, a₁₃ = 1×2 + 3×4 = 14, a₂₁ = -1×1 + 1×3 = 2, a₂₂ = -1×(-1) + 1×1 = 2, a₂₃ = -1×2 + 1×4 = 2, a₃₁ = 2×1 + 4×3 = 14, a₃₂ = 2×(-1) + 4×1 = 2, a₃₃ = 2×2 + 4×4 = 20

$A^{T} A=\left[\begin{array}{ccc}10&2&14\\2&2&2\\14&2&20\end{array}\right]$

FE = $\left[\begin{array}{ccc}i&1-3i\\0&4+i\end{array}\right]$ $\left[\begin{array}{ccc}2-i&1+i\\-i&2+4i\end{array}\right]$

a₁₁ = i×(2-i) + (1-3i)×(-i) = -2+i, a₁₂ = i×(1+i) + (1-3i)×(2+4i) = 13-i, a₂₁ = 0×(2-i) + (4+i)×(-i) = 1-4i, a₂₂ = 0×(1+i) + (4+i)×(2+4i) = 4+18i

FE = $\left[\begin{array}{ccc}-2+i&13-i\\1-4i&4+18i\end{array}\right]$

Hence we can obtain the results of the required matrices using matrix multiplication.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: Let A = $\left[\begin{array}{ccc}1&-1&2\\3&1&4\end{array}\right]$ , B = $\left[\begin{array}{ccc}2&-1&3\\5&1&2\\4&6&-2\end{array}\right]$ , C = $\left[\begin{array}{ccc}1\\-1\\2\end{array}\right]$ , D = $\left[\begin{array}{ccc}2&-2&3\end{array}\right]$ , E = $\left[\begin{array}{ccc}2-i&1+i\\-i&2+4i\end{array}\right]$ , F = $\left[\begin{array}{ccc}i&1-3i\\0&4+i\end{array}\right]$

For each item, decide whether or not the given expression is defined. for each item that is defined, compute the result.

AB, BC, CA, CD, $C^{T} A^{T}$ , F², $BD^{T}$ , $A^{T} A$ and FE

Learn more about matrix here:

brainly.com/question/28180105

#SPJ4

8 0

2 years ago

1) Find the area of  ABC to the nearest square centimeter.

satela [25.4K]

that is right correct answer

3 0

3 years ago