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3 years ago

May yall help me what the answer????

Mathematics

2 answers:

Elena-2011 [213]3 years ago

7 0

Answer: i have no idea sowy

Step-by-step explanation:

evablogger [386]3 years ago

4 0

Answer:

i have no idea too- srry.. good luck

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Philip buys the pizza shown below: A circular pizza is shown. The center is labeled as point A and a point on the circumference

Mnenie [13.5K]

So, the line created from the center of the circle to the edge of the circle creates the radius. If it were a straight line from one end of the circle, through the center, to the other end, that would create the diameter.

So, the answer is radius

8 0

4 years ago

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dane used a ruler and a protractor to draw a rhombus with a side of 4 cm inlength and an angle that measures 110 degrees choose

lara [203]

Answer:

70 °

Step-by-step explanation:

We have that one side of the rhombus is equal to 4 cm and an angle of the rhombus is 110 degrees. We know that the adjacent sides of a rhombus are supplementary, which means that the sum of the angles must be 180 degrees.

We can assume that the other angle is x.

110 ° + x = 180 °

x = 180 ° -110 °

x = 70 °

The measure of the missing angle on the diagram will be 70 °.

3 0

3 years ago

Which best describes the resulting three-dimensional

Harman [31]

Answer: C

Step-by-step explanation:

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3 years ago

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The probability that a lab specimen contains high levels of contamination is 0.10. A group of 5 independent samples are checked.

zhannawk [14.2K]

Answer with Step-by-step explanation:

In case of Bernoulli trails

The probability that a random variable occurs 'r' times in 'n' trails is given by

$P(E)=\binom{n}{r}p^r(1-p)^{n-r}$

where

'p' is the probability of success of the event

Part a)

probability that no contamination occurs can be found by putting r = 0

Thus we get

$P(E_1)=\binom{5}{0}0.1^0(1-0.1)^{5}=0.5905$

part b)

The probability that at least 1 contamination occurs is given by

$P(E)=1-(1-p)^{n}$

Applying values we get

$P(E_2)=1-(1-0.1)^{5}=0.4096$

3 0

3 years ago

Determine whether the following matrices are inverse matrices of one another

Liono4ka [1.6K]

9514 1404 393

Answer:

they are not

Step-by-step explanation:

The inverse matrix is the transpose of the cofactor matrix, divided by the determinant.

The cofactor matrix for a 2×2 matrix is ...

$\left[\begin{array}{cc}a_{22}&-a_{21}\\-a_{12}&a_{11}\end{array}\right]$

The transpose of this will have the off-diagonal terms swapped, so the inverse matrix is ...

$\displaystyle\frac{\left[\begin{array}{cc}a_{22}&-a_{12}\\-a_{21}&a_{11}\end{array}\right]}{a_{11}a_{22}-a_{21}a_{12}}$

We see that the second matrix is the transpose of the cofactor matrix, but the determinant is (5)(2)-(3)(4) = -2, so there has clearly been no division by the determinant. The actual inverse matrix of the first one shown is ...

$\left[\begin{array}{cc}-1&2\\1.5&-2.5\end{array}\right]$

_____

You can compute the matrix product to see if you get an identity matrix. Here, the upper left term in the product is ...

(5)(2) +(4)(-3) = -2 . . . . . not 1, so the product matrix is not an identity matrix

The matrices are not inverses.

5 0

3 years ago