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3 years ago

HELP PLS DUE TOMORROW WILL GIVE BRAINLIEST

Mathematics

1 answer:

Scilla [17]3 years ago

5 0

Answer:

SSS: LM = AB and MN = AC

HL: AC = MN

Step-by-step explanation:

✔️Information that is necessary to prove that both triangles are congruent by SSS are:

LM = AB

MN = AC

This will is because all three corresponding sides must be congruent to each other to satisfy the SSS Congruence criterion.

✔️Information necessary to prove that the triangles are congruent by HL:

We already know that one of the corresponding legs are congruent to each either. The additional information necessary to satisfy the HL Congruence criterion is to know the hypotenuse length of each which must be congruent to each other. This, information necessary would be:

AC = MN

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What is the slope of a line that is parallel to the line with the following equation?

Natali [406]

Answer:

Step-by-step explanation:

we know that,

Parallel lines have the same slope

given equation is in the form given by

$\displaystyle y = mx + b$

where m is know

as $m_{parallel}=m$ , the answer is -⅔

3 0

3 years ago

The resting heart rate for an adult horse should average about µ = 47 beats per minute with a (95% of data) range from 19 to 75

KatRina [158]

Answer:

a. 0.0582 = 5.82% probability that the heart rate is less than 25 beats per minute.

b. 0.1762 = 17.62% probability that the heart rate is greater than 60 beats per minute.

c. 0.7656 = 76.56% probability that the heart rate is between 25 and 60 beats per minute

Step-by-step explanation:

Empirical Rule:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean:

$\mu = 47$

(95% of data) range from 19 to 75 beats per minute.

This means that between 19 and 75, by the Empirical Rule, there are 4 standard deviations. So

$4\sigma = 75 - 19$

$4\sigma = 56$

$\sigma = \frac{56}{4} = 14$

a. What is the probability that the heart rate is less than 25 beats per minute?

This is the p-value of Z when X = 25. So

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

$Z = \frac{25 - 47}{14}$

$Z = -1.57$

$Z = -1.57$ has a p-value of 0.0582.

0.0582 = 5.82% probability that the heart rate is less than 25 beats per minute.

b. What is the probability that the heart rate is greater than 60 beats per minute?

This is 1 subtracted by the p-value of Z when X = 60. So

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

$Z = \frac{60 - 47}{14}$

$Z = 0.93$

$Z = 0.93$ has a p-value of 0.8238.

1 - 0.8238 = 0.1762

0.1762 = 17.62% probability that the heart rate is greater than 60 beats per minute.

c. What is the probability that the heart rate is between 25 and 60 beats per minute?

This is the p-value of Z when X = 60 subtracted by the p-value of Z when X = 25. From the previous two items, we have these two p-values. So

0.8238 - 0.0582 = 0.7656

0.7656 = 76.56% probability that the heart rate is between 25 and 60 beats per minute

3 0

3 years ago

Use a surface integral to ﬁnd the surface area of the portion of the sphere xUse a surface integral to ﬁnd the surface area of t

brilliants [131]

Parameterize this surface (call it $S$ ) by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\dfrac\pi4$ . The limits on $u$ should be obvious. We find the upper limit for $v$ by solving for $v$ where the sphere and cone intersect:

$\begin{cases}x^2+y^2+z^2=1\\z=\sqrt{x^2+y^2}\end{cases}\implies x^2+y^2=\dfrac12$

$\implies(\cos u\sin v)^2+(\sin u\sin v)^2=\dfrac12$

$\implies\sin^2v=\dfrac12$

$\implies\sin v=\dfrac1{\sqrt2}\implies v=\dfrac\pi4$

Take the normal vector to $S$ to be

$\vec r_u\times\vec r_v=-\cos u\sin^2v\,\vec\imath-\sin u\sin^2v\,\vec\jmath-\cos v\sin v\,\vec k$

(orientation does not matter here)

Then the area of $S$ is

$\displaystyle\iint_S\mathrm dA=\iint_S\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{\pi/4}\int_0^{2\pi}\sin v\,\mathrm du\,\mathrm dv$

$=\displaystyle2\pi\int_0^{\pi/4}\sin v\,\mathrm dv=\boxed{(2-\sqrt2)\pi}$

4 0

3 years ago

7. Consider the purchase of two cereal boxes.

AveGali [126]

Answer:

a) 0.25

b) 0.25

c) 0.0625

Step-by-step explanation:

The complete question is:

Do you remember when breakfast cereal companies placed prizes in boxes of cereal? Possibly you recall that when a certain prize or toy was particularly special to children, it increased their interest in trying to get that toy. How many boxes of cereal would a customer have to buy to get that toy? Companies used this strategy to sell their cereal.

One of these companies put one of the following toys in its cereal boxes: a block (B), a toy watch (W), a toy ring (R), and a toy airplane (A). A machine that placed the toy in the box was programmed to select a toy by drawing a random number of 1 to 4. If a 1 was selected, the block (or B) was placed in the box; if a 2 was selected, a watch (or W) was placed in the box; if a 3 was selected, a ring (or R) was placed in the box; and if a 4 was selected, an airplane (or A) was placed in the box. When this promotion was launched, young children were especially interested in getting the toy airplane.

What is the probability of getting an airplane in the first cereal box?

Since the machine randomly selects toys, each toy has the same probability of being obtained in a cereal box.

Then, the total outcomes are 4 and the probability of getting an airplane in the first cereal box is 0.25 (25%).

What is the probability of getting an airplane in the second cereal box?

Two independent events do not change the probability of occurrence of one event or another.

The probability of getting an airplane in the second cereal box is 0.25 (25%).

What is the probability of getting airplanes in both cereal boxes?

P(1°∩2°)= P(1°) × P(2°) = $\frac{1}{4} \times \frac{1}{4} =\frac{1}{16}$

P(1°∩2°)= 0.0625 = 6.25%

4 0

3 years ago

Please help. Which is the largest?

sergey [27]

Answer:

the first one

Step-by-step explanation:

5 0

3 years ago