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Mars2501 [29]
3 years ago
12

The coordinates of the vertices of △JKL are J(−2, 1) , K(−1, 3) , and L(−3, 4) . The coordinates of the vertices of △J′K′L′ are

J′(−2, −1) , K′(−1, −3) , and L′(−3, −3) .
Which statement correctly describes the relationship between △JKL and △J′K′L′?


A) △JKL is congruent to △J′K′L′ because you can map △JKL to △J′K′L′ using a reflection across the y-axis, which is a rigid motion.

B) △JKL is congruent to △J′K′L′ because you can map △JKL to △J′K′L′ using a translation 2 units down, which is a rigid motion.

C) △JKL is not congruent to △J′K′L′ because there is no sequence of rigid motions that maps △JKL to △J′K′L′ .

D) △JKL is congruent to △J′K′L′ because you can map △JKL to △J′K′L′ using a reflection across the x-axis, which is a rigid motion.

Mathematics
2 answers:
Nadusha1986 [10]3 years ago
7 0
The answer is C) △JKL is not congruent to △J′K′L′ because there is no sequence of rigid motions that maps △JKL to △J′K′L′ .
Gemiola [76]3 years ago
3 0

Answer:

C) △JKL is not congruent to △J′K′L′ because there is no sequence of rigid motions that maps △JKL to △J′K′L′.

Step-by-step explanation:

If L' were (-3,-4), it would be a reflection of L across the x-axis as J' and K' are with respect to J and K. Unfortunately, because it is not, the side lengths J'L' and K'L' of triangle J'K'L' are different from those of triangle JKL. This ensures the triangles JKL and J'K'L' are <em>not congruent</em>.

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2. The Welcher Adult Intelligence Test Scale is composed of a number of subtests. On one subtest, the raw scores have a mean of
IgorC [24]

Answer:

a) 37.31 b) 42.70 c) 0.57 d) 0.09

Step-by-step explaanation:

We are regarding a normal distribution with a mean of 35 and a standard deviation of 6, i.e., \mu = 35 and \sigma = 6. We know that the probability density function for a normal distribution with a mean of \mu and a standard deviation of \sigma is given by

f(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp[-\frac{(x-\mu)^{2}}{2\sigma^{2}}]

in this case we have

f(x) = \frac{1}{\sqrt{2\pi}6}\exp[-\frac{(x-35)^{2}}{2(6^{2})}]

Let X be the random variable that represents a row score, we find the values we are seeking in the following way

a)  we are looking for a number x_{0} such that

P(X\leq x_{0}) = \int\limits^{x_{0}}_{-\infty} {f(x)} \, dx = 0.65, this number is x_{0}=37.31

you can find this answer using the R statistical programming languange and the instruction qnorm(0.65, mean = 35, sd = 6)

b) we are looking for a number  x_{1} such that

P(X\leq x_{1}) = \int\limits^{x_{1}}_{-\infty} {f(x)} \, dx = 0.9, this number is x_{1}=42.70

you can find this answer using the R statistical programming languange and the instruction qnorm(0.9, mean = 35, sd = 6)

c) we find this probability as

P(28\leq X\leq 38)=\int\limits^{38}_{28} {f(x)} \, dx = 0.57

you can find this answer using the R statistical programming languange and the instruction pnorm(38, mean = 35, sd = 6) -pnorm(28, mean = 35, sd = 6)

d) we find this probability as

P(41\leq X\leq 44)=\int\limits^{44}_{41} {f(x)} \, dx = 0.09

you can find this answer using the R statistical programming languange and the instruction pnorm(44, mean = 35, sd = 6) -pnorm(41, mean = 35, sd = 6)

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3 years ago
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statuscvo [17]
Let's start by visualising this concept.

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We can see that it starts to form a geometric sequence, with the common ratio being 2.

For the first question, we simply want the fifteenth term, so we just use the nth term geometric form:
T_n = ar^{n - 1}
T_{15} = 2^{14} = 16384

Thus, there are 16, 384 grains on the fifteenth square.

The second question begs the same process, only this time, it's a summation. Using our sum to n terms of geometric sequence, we get:
S_n = \frac{a(r^{n} - 1)}{r - 1}
S_{15} = \frac{2^{15} - 1}{2 - 1}
S_{15} = 2^{15} - 1 = 32767

Thus, there are 32, 767 total grains on the first 15 squares, and you should be able to work the rest from here.
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Step-by-step explanation:

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