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Montano1993 [528]
3 years ago
15

Can someone solve this

Mathematics
1 answer:
Natalka [10]3 years ago
4 0

Option D:

ΔCAN ≅ ΔWNA by SAS congruence rule.

Solution:

Given data:

m∠CNA = m∠WAN and CN = WA

To prove that ΔCAN ≅ ΔWNA:

In ΔCAN and ΔWNA,

CN = WA (given side)

∠CNA = ∠WAN (given angle)

NA = NA (reflexive side)

Therefore, ΔCAN ≅ ΔWNA by SAS congruence rule.

Hence option D is the correct answer.

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The enrollment at Temple Junior High School has a change of ─ 60 over a 3 year period. What is the average change per year?
GaryK [48]

Answer:

- 20

Step-by-step explanation:

Total Change in enrollment = - 60

Period of time at which change occurred = 3 years

Average change per year :

Total change in enrollment / period at which change occurred

= - 60 / 3

= - 20

Average change of - 20 enrollments per year

4 0
2 years ago
It is important that face masks used by firefighters be able to withstand high temperatures because firefighters commonly work i
USPshnik [31]

Answer:

The 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574). This means that we are 93% sure that the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 55, \pi = \frac{24}{55} = 0.4364

93% confidence level

So \alpha = 0.07, z is the value of Z that has a pvalue of 1 - \frac{0.07}{2} = 0.965, so Z = 1.81.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4364 - 1.81\sqrt{\frac{0.4364*0.5636}{55}} = 0.3154

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4364 + 1.81\sqrt{\frac{0.4364*0.5636}{55}} = 0.5574

The 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574). This means that we are 93% sure that the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574).

8 0
3 years ago
Helpppp i literally have no idea which one it is
SVETLANKA909090 [29]

Ya what the person above me said

7 0
2 years ago
What are the special right triangle measures?
erik [133]
The hypotenuse is 2x, the base is x*sqrt(3) and the left base if x
5 0
2 years ago
Earth revolves around the Sun in 365 days, but the planet Mercury does so in only 88 days. Compared to Earth, how many more comp
Elena L [17]

Answer:

In a 1000 days Mercury will make 8.624 more revolutions than Earth!

Step-by-step explanation:

<u>Earth</u> takes 365 days to make a complete revolution.

we can say that the rotational speed of Earth is \dfrac{1}{365}\frac{\text{rev}}{\text{day}}

so, in a 1000 days how many revolutions will Earth have made?

\dfrac{1}{365}\frac{\text{rev}}{\text{day}}\times(1000\,\text{day})

see that the 'day's cancel out. Simply multiply the two numbers!

2.739\text{rev}

<u>Next, do the same for Mercury</u>

we can say that the rotational speed of Mercury is \dfrac{1}{88}\frac{\text{rev}}{\text{day}}

as it completes 1 revolution in 88 days.

so, in a 1000 days how many revolutions will Mercury have made?

\dfrac{1}{88}\frac{\text{rev}}{\text{day}}\times(1000\,\text{day})

multiplying by 1/88 by 1000 yields:  

11.363\text{rev}

The Question is how many MORE complete revolutions will Mercury make in 1000 days?

We need to subtract the revolutions of Mercury with Earth to find how far ahead is Mercury from Earth

11.363\text{rev} - 2.739\text{rev}

8.624\text{rev}

In a 1000 days Mercury will make 8.624 more revolutions than Earth!

5 0
3 years ago
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