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1 year ago

Using the transformation T: (x, y) (x + 2, y + 1), find the distance named.

Mathematics

2 answers:

konstantin123 [22]1 year ago

8 0

Answer:

The distance CC' is $\sqrt5units$

Step-by-step explanation:

Given the transformation T: (x, y) (x + 2, y + 1)

we have to find the distance CC'

Let coordinate of C are (a,b).

Now, by using transformation T the coordinates of C' are (a+2,a+1)

By using distance formula,

$CC'=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\= \sqrt{(a+2-a)^2+(b+1-b)^2}\\\\=\sqrt{4+1}=\sqrt5 units$

Hence, the distance CC' is $\sqrt5units$

stellarik [79]1 year ago

3 0

I hope this helps you

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Write a different expression that has the same quotient as -3/8 and 3/-8.

matrenka [14]

An equivalent fraction is a fraction formed by multiplying the numerator and the denominator of a fraction by a common number.

$-\frac{3}{8} =- \frac{3\times2}{8\times2} = -\frac{6}{16}$

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1 year ago

Which expression is equivalent to StartRoot 128 x Superscript 8 Baseline y cubed z Superscript 9 Baseline EndRoot? Assume y grea

Nataly [62]

Answer:

Option C.

Step-by-step explanation:

The given expression is

$\sqrt{128x^8y^3z^9}$

It can be written as

$\sqrt{(64\cdot 2)\cdot x^8\cdot y^{2+1}\cdot z^{8+1}}$

$=\sqrt{8^2\cdot 2\cdot (x^4)^2\cdot y^{2}\cdot y\cdot (z^{4})^2\cdot z}$

$=\sqrt{(8x^4yz^4)^2\cdot 2yz}$

$=8x^4yz^4\sqrt{2yz}$

Therefore, the correct option is C.

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1 year ago

Two professors are applying for grants. Professor Jane has a probability of 0.61 of being funded. Professor Joe has probability

Anna11 [10]

Answer:

a. 0.1647;
b. 0.7153;
c. 0.4453.

Step-by-step explanation:

By definition, if two event $A$ and $B$ are independent, then

$P(A\cap B) = P(A) \cdot P(B)$ . ( $P(A\cap B)$ is the probability that the outcome of both event $A$ and event $B$ are true.)

Since the outcome of these two events are independent,

$\begin{aligned}P(\texttt{Jane} \cap \texttt{Joe}) &= P(\texttt{Jane}) \cdot P(\texttt{Joe})\\ &= 0.61 \times 0.27 \\&= 0.1647\end{aligned}$ .

The logic not operator $\lnot$ or the prime superscript $^{\prime}$ denotes that an event does not happen.

$P(\texttt{Jane}^{\prime}) = 1 - P(\texttt{Jane}) = 1- 0.61 = 0.39$ .

$P(\texttt{Joe}^{\prime}) = 1 - P(\texttt{Joe}) = 1- 0.27 = 0.73$ .

Since the two events $\texttt{Jane}$ and $\texttt{Joe}$ , $\texttt{Jane}^{\prime}$ and $\texttt{Joe}^{\prime}$ are also independent. Probability that neither professor got funded:

$P(\texttt{Jane}^{\prime} \cap \texttt{Joe}^{\prime}) = P(\texttt{Jane}^{\prime}) \cdot P(\texttt{Joe}^{\prime}) = 0.39 \times 0.73 = 0.2847$ .

Probability that at least one professor got funded- in other words, it is not true that neither professor got funded:

$P((\texttt{Jane}^{\prime} \cap \texttt{Joe}^{\prime})^{\prime}) = 1- P(\texttt{Jane}^{\prime} \cap \texttt{Joe}^{\prime}) = 0.7153$ .

Similarly, since the two events $\texttt{Jane}$ and $\texttt{Joe}$ , $\texttt{Jane}$ and $\texttt{Joe}^{\prime}$ are also independent. Probability that Jane but not Joe got funded:

$P(\texttt{Jane} \cap (\texttt{Joe}^{\prime})) = P(\texttt{Jane}) \cdot P(\texttt{Joe}^{\prime}) = 0.61 \times 0.73 = 0.4453$ .

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1 year ago

The decimal expansion of 141/120 will terminate after how many places of decimals

zmey [24]

Answer:

nixgaa

Step-by-step explanation:

brih

4 0

1 year ago

Food allergies affect an estimated 7% of children under age 5 in the US. What is the probability that in a kindergarden class of

bezimeni [28]

Answer:

Probability of atleast one of 12 student has food allergies ≈ 0.58 ( approx)

Step-by-step explanation:

Given: Probability of a children under age 5 has food allergies = 7%

= $\frac{7}{100}$

To find : Probability of atleast one of 12 student has food allergies

Probability of a chindren under age 5 does not have food allergies = $1-\frac{7}{100}$

⇒ prob = $\frac{93}{100}$

now we find Probability of atleast one of 12 student has food allergies this means we have to find prob of 1 student, 2 student, 3 student, till 12 student have allergy out of 12 student of class then add all prob.

But instead of finding all these probability we find probability of student having no allergy.i.e., 0 student then subtract it from 1(total probability)

Probability of 0 student having allergy out of 12 student = $(\frac{93}{100})^{12}$

Therefore, Probability of atleast one of 12 student has food allergies

= $1-(\frac{93}{100})^{12}$

= $0.581403702521$

≈ 0.58 ( approx)

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1 year ago

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