All categories

3 years ago

Anyone think they could help me with this question???

Mathematics

1 answer:

Slav-nsk [51]3 years ago

3 0

Answer:

Step-by-step explanation:

As we can see the triangles are congruent so

< F = < C ( being corresponding angles of congruent triangle)

< C = 29

Now

<A +<B + <C = 180° { being sum of angles of triangle }

5x + 1 + 90 + 29 = 180°

5x = 180° - 120°

5x = 60°

x = 12°

<A = 5 * 12° + 1 = 61°

Hope it will help :)

You might be interested in

Please help. I suck at geometry

Natasha_Volkova [10]

Answer:

Complementary angle is 90 degree and supplementary angle is 180 degree

8 0

3 years ago

HELP PLEASE!!!!!!!!!!!!!!!!!!

Doss [256]

I can not see the answer or the question

6 0

3 years ago

Use the line plot above to answer all 4 questions. Label your responses (a), (b), (c), and (d). When asked to explain you must u

mojhsa [17]

A) it is left skewed
B) the median is 5
C) the mean is 5.15
D) the mean would be more affected (a change of 1.05 versus a change of 0.5).

The majority of the data is to the right of the graph; this means it is left skewed.

To find the median, write all of the data values out:
2, 3, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7

The middle value is 5.

We find the sum of this set of values and divide by 13, the number of data points, to find the mean:
2+3+4+4+5+5+5+6+6+6+7+7+7 = 67/13 = 5.15

If we added an additional data value at 20, the new median would be 5.5. The new mean would be (67+20)/14 = 6.2. The mean changes more than the median.

4 0

3 years ago

n a call center the number of received calls in a day can be modeledby a Poisson random variable. We know that, on average, 0.5%

guapka [62]

Answer:Pois(ln(200))

Step-byy-step explanation:

Let N be the number of received calls in a day

That is

N∼Pois(λ).

0.5% = 0.5/100 = 1/200 of no calls

P(N=0)=e^−λ=1/200

-λ=e^(1/200)

λ=in(200)

Our number of calls in a day is distributed Pois(ln(200)).

7 0

4 years ago

Read 2 more answers

Intellectual development (Perry) scores were determined for 21 students in a first-year, project-based design course. (Recall th

Anit [1.1K]

Answer:

The 99% confidence interval is (3.0493, 3.4907).

We are 99% sure that the true mean of the students Perry score is in the above interval.

Step-by-step explanation:

Our sample size is 21.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 21-1 = 20$ .

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.99}{2} = \frac{0.01}{2} = 0.005$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 20 and 0.005 in the two-sided t-distribution table, we have $T = 2.528$