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4 years ago

Question 20 points if correct

Mathematics

2 answers:

True [87]4 years ago

7 0

Is This On Khan Academy because I be needing help

Vlada [557]4 years ago

5 0

Answer:

B = 155° , C = 25° , D = 155°

Step-by-step explanation:

Given

A = 25°

C = A = 25° [ Vertically Opposite angles ]

Now

A + B = 180°( being linear pair)

25° + B = 180°

B = 180° - 25°

B = 155°

Also

D = B = 155°( vertically opposite angles )

You might be interested in

Over what interval is the function differentiable? Y=3/2(x-1)

jolli1 [7]

Answer-

The interval where the function is diffrentiable is

$[-\infty,1)\ \bigcup\ (1,\infty]$

Solution-

The given expression is,

$y=\dfrac{3}{2(x-1)}$

The function will be differentiable where it is continuous and it will not be differentiable, where the function is not continuous.

The function continuous everywhere except at x = 1, because

$\Rightarrow 2(x-1)=0\\\\\Rightarrow x-1=0\\\\\Rightarrow x=1$

at x = 1, its limit does not exist.

Therefore, apart from x=1, this function is differentiable everywhere. The interval will be

$[-\infty,1)\ \bigcup\ (1,\infty]$

3 0

3 years ago

4. Ariel loves to go to baseball games. This season she has already attended 23 New York Ducks games. If

san4es73 [151]

Answer: She attended to 7 weekday games.

Step-by-step explanation:

Let's denote our variables as:

Wd = number of weekday games that she attended.

We = number of weekend games that she attended.

We know that in total, she attended to 23 games, so we have:

Wd + We = 23.

"If Ariel attended nine more weekend games than weekday games" can be written as:

We = Wd + 9.

Then we have the system of equations:

Wd + We = 23.

We = Wd + 9.

The first step would be to replace the second equation into the first one, as:

Wd + We = 23.

Wd + (Wd + 9) = 23

Now we can solve this for Wd.

2*Wd + 9 = 23

2*Wd = 23 - 9 = 14

Wd = 14/2 = 7

She attended to 7 weekday games.

And:

We + 7 = 23

We = 23 - 7 = 16

She attended at 16 weekend games.

6 0

4 years ago

According to a poll, 76% of California adults (385 out of 506 surveyed) feel that education is one of the top issues facing Cali

lina2011 [118]

Answer:

The error bound is 3.125%.

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 interval $1-\alpha$ , 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:

A sample of 506 California adults.. This means that $n = 506$ .

76% of California adults (385 out of 506 surveyed) feel that education is one of the top issues facing California. This means that $\pi = 0.76$

We wish to construct a 90% confidence interval

So $\alpha = 0.10$ , z is the value of Z that has a pvalue of $1 - \frac{0.10}{2} = 0.95$ , so $z = 1.645$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.76 - 1.645\sqrt{\frac{0.76*0.24}{506}} = 0.7288$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.76 + 1.645\sqrt{\frac{0.76*0.24}{506}} = 0.7913$

The error bound of the confidence interval is the division by 2 of the subtraction of the upper limit by the lower limit. So:

$EBM = \frac{0.7913 - 0.7288}{2} = 0.03125$

The error bound is 3.125%.

3 0

3 years ago

The ages of 3 siblings combined is 27.the oldest is twice the age of the youngest.the middle child is 3 years older than the you

Olenka [21]

I ended up with the ages of 12, 3, and 2
if u multiply 12 by 2 u end up with 24 then u add 3 and get 27
but I think i have i worked out wrong

8 0

4 years ago

The estimated probability of a football player scoring a touchdown during a particular game is 39%. Of several simulations of th

VLD [36.1K]

Answer:

The percentage of the simulations of two games that the football player would most likely get a touchdown in each of the two games = 15.21%

Step-by-step explanation:

The probability of scoring a touchdown in one game for the football player = 39% = 0.39

Probability of scoring a touchdown in each of the two consecutive games = (Probability of scoring a touchdown in the first game) × (Probability of scoring a touchdown in the second game)

Since the probability of scoring a touchdown in each game the football player plays in is independent of one another.

Probability of scoring a touchdown in the first game = Probability of scoring a touchdown in the second game = 0.39

Probability of scoring a touchdown in each of the two consecutive games = 0.39 × 0.39 = 0.1521

If 100 simulations of the two games are made, the player scores in each of the two games in 15.21 of them, hence, the percentage of the simulations that the football player would most likely get a touchdown in each of the two games = 15.21%

Hope this Helps!!

4 0

4 years ago