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2 years ago

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I(5, 3), J(5, -3), and L(-4, 3) are three vertices of rectangle IJKL. What are the coordinates of the fourth vertex, K, of recta

ngle IJKL? A. (-4, -3) B. (5, 3) C. (5, -3) D. (-4, 3)

2 answers:

Ad libitum [116K]2 years ago

6 0

Answer:

A. $K(x,y) = (-4,-3)$

Step-by-step explanation:

On a cartesian plane, rectangle whose each pair of parallel side are parallel to respective axis is a quadrilateral must observe the following conditions:

$I(x,y) = (a, b)$ , $J(x,y) = (a, c)$ , $L(x,y) = (d, b)$ , $K(x,y) = (d, c)$

If we know that $a = 5$ , $b = 3$ , $c = -3$ , $d = -4$ , then the coordinates of the vertex K of the rectangle IJKL is:

$K(x,y) = (-4,-3)$

Hence, the correct answer is A.

Mice21 [21]2 years ago

6 0

Answer:A. -4 -3

Step-by-step explanation:

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Mrs. Olga’s $250 investment earns 19% interest compounded annually. What will her balance be after 3 years?

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Answer:

The answer is A. $421.29

Step-by-step explanation:

Just keep finding 19% of the numbers next, and you add that all up and you get the answer.

250 x 19%=47.5. (250+47.5)=> 297.5x19%= 56.525. (297.5+56.525) =>354.025x19%=67.26475

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Between 0 degrees Celsius and 30 degrees Celsius, the volume V (in cubic centimeters) of 1 kg of water at a temperature T is giv

Ira Lisetskai [31]

Answer:

T = 3.967 C

Step-by-step explanation:

Density = mass / volume

Use the mass = 1kg and volume as the equation given V, we will come up with the following equation

D = 1 / 999.87−0.06426T+0.0085043T^2−0.0000679T^3

= (999.87−0.06426T+0.0085043T^2−0.0000679T^3)^-1

Find the first derivative of D with respect to temperature T

dD/dT = $\dfrac{70000000\left(291T^2-24298T+91800\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^2}$

Let dD/dT = 0 to find the critical value we will get

$\dfrac{70000000\left(291T^2-24298T+91800\right)}$ = 0

Using formula of quadratic, we get the roots:

T = 79.53 and T = 3.967

Since the temperature is only between 0 and 30, pick T = 3.967

Find 2nd derivative to check whether the equation will have maximum value:

$-\dfrac{140000000\left(395178T^4-65993368T^3+3286558821T^2+2886200857800T-121415215620000\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^3}$

Substituting the value with T=3.967,

d2D/dT2 = -1.54 x 10^(-8) a negative value. Hence It is a maximum value

Substitute T =3.967 into equation V, we get V = 0.001 i.e. the volume when the the density is the highest is at 0.001 m3 with density of

D = 1/0.001 = 1000 kg/m3

Therefore T = 3.967 C

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2 years ago

T varies inversely as the square root of u.<br>t = 3 when u = 4.<br>Find t when u = 49.

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Answer:

t = $\frac{6}{7}$

Step-by-step explanation:

Given that t varies inversely as the square root of u then the equation relating them is

t = $\frac{k}{\sqrt{u} }$ ← k is the constant of variation

To find k use the condition t = 3 when u = 4, thus

k = t $\sqrt{u}$ = 3 × $\sqrt{4}$ = 3 × 2 = 6

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In choice situations of this type, subjects often exhibit the "center stage effect," which is a tendency to choose the item in t

victus00 [196]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The probability that he or she would choose the pair of socks in the center position is $p =\frac{1}{5}$

The correct answer choice is

X has a binomial distribution with parameters n=100 and p=1/5

b

The mean is $\mu = 20$

The standard deviation is $\sigma=4$

c

The probability, $P =0.0002$

d

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

Using the R the probability $Pe = 0.0003$

The probabilities $P \approx Pe$

Step-by-step explanation:

Since the person selects his or her desired pair of socks at random , then the probability that the person would choose the pair of socks in the center position from all the five identical pair is mathematically evaluated as

$p =\frac{1}{5}$

$=0.2$

The mean of this distribution is mathematical represented as

$\mu = np$

substituting the value

$\mu = 100 * 0.2$

$\mu = 20$

The standard deviation is mathematically represented as

$\sigma = \sqrt{np (1-p)}$

substituting the value

$= \sqrt{100 * 0,2 (1-0.2)}$

$\sigma=4$

Applying normal approximation the probability that 34 or more subjects would choose the item in the center if each subject were selecting his or her preferred pair of socks at random would be mathematically represented as

$P=P(X \ge 34 )$

By standardizing the normal approximation we have that

$P(X \ge 34) \approx P(Z \ge z)$

Now z is mathematically evaluated as

$z = \frac{x-\mu}{\sigma }$

Substituting values

$z = \frac{34-20}{4}$

$=3.5$

So using the z table the $P(Z \ge 3.5)$ is 0.0002

The probability P and Pe that 34 or more subject would choose the center pair is very small So

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

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2 years ago