For the pair of similar triangles, find the value of x.

I don't know how to do this

Phantasy [73]

I completely agree with the above statement

3 0

3 years ago

The product of a constant factor 15 and a 2-term factor x+4

dezoksy [38]

The product of 15 and $x+4$ is $15(x+4) = 15x+60$ . This is a linear binomial.

$4x^3-2x-9-(2x^3)$ is a cubic trinomial. In your original question, I'm not sure what the $5x+3$ means.

6 0

2 years ago

84,552 rounded to the nearest 1000 and 100 and 10

Juli2301 [7.4K]

Answer:

85,000; 84,600; 84,550

Step-by-step explanation:

84,552 is closer to 85,000

84,552 is closer to 84,600

84,552 is closer to 84,550

7 0

3 years ago

Can someone produce SPSS results by including the the intercept (a.k.a. the constant), standard error of estimate, unstandardize

Firlakuza [10]

Answer:these are all odd

Step-by-step explanation:

the p-value for the unstandardized coefficient, R-square, Beta (i.e. standardized regression coefficient) using Excel with this given data?

Overall: 3.4, 2.9, 2.6, 3.8, 3.0, 2.5, 3.9, 4.3, 3.8, 3.4, 2.8, 2.9, 4.1, 2.7, 3.9, 4.1, 4.2, 3.1, 4.1, 3.6, 4.3, 4.0, 2.1, 3.8, 2.7, 4.4, 3.1, 3.6, 3.9, 2.9, 3.7, 2.8, 3.3, 3.7, 4.2, 2.9, 3.9, 3.5, 3.8, 4.0, 3.1, 4.2, 2.9, 3.9, 3.5, 3.8, 4.0, 3.1, 4.2, 3.0, 4.8, 3.0, 4.4, 4.4, 3.4, 4.0, 3.5

Teach: 3.8, 2.8, 2.2, 3.5, 3.2, 2.7, 4.1, 4.2, 3.7, 3.7, 3.3, 3.3, 4.1, 3.1, 2.9, 4.5, 4.3, 3.7, 4.2, 4.0, 3.7, 4.0 2.9, 4.0, 3.3, 4.4, 3.4, 3.8, 3.7, 3.1, 3.8, 3.2, 3.5, 3.8, 4.4, 3.7, 4.00, 3.4, 3.2, 3.8, 3.7, 4.3, 3.4, 4.0 3.1, 4.5, 4.8, 3.4, 4.2, 3.4

Exam: 3.8, 3.2, 1.9, 3.5, 2.8, 3.8, 3.8, 4.1, 3.6, 3.6, 3.5, 3.3, 3.6, 3.8, 3.8, 4.2, 4.1, 4.0, 4.3, 4.2, 4.0, 4.1, 2.7, 4.4, 4.4, 4.3, 3.6, 4.1, 4.2, 3.6, 4.4, 3.4, 3.2, 3.7, 4.3, 4.1, 3.7, 4.0, 3.6, 4.0, 3.7, 4.2, 4.2, 4.1, 3.2, 4.5, 4.3, 3.6, 4.00, 3.9

Knowledge: 4.5, 3.8, 3.9, 4.1, 3.5, 4.2, 4.5, 4.7, 4.1, 4.1, 3.9, 3.9, 4.0, 4.1, 4.5, 4.5, 4.5, 4.5, 4.7, 4.0, 4.5, 4.6, 3.7, 4.1, 3.6, 4.4, 3.3, 3.8, 4.2, 3.8, 4.0, 3.1, 4.4, 4.3, 5.00 4.2, 4.5, 4.5, 4.7, 4.3, 4.00, 4.2, 3.8, 4.9, 3.7, 4.6, 4.3, 3.5, 4.4, 4.4

Grade: 3.5, 3.2, 2.8, 3.3,

6 0

2 years ago

Assume that the Poisson distribution applies and that the mean number of aircraft accidents is 9 per month. Find P(0), the prob

Ugo [173]

Answer: 0.0001

It is unlikely to have a month with no aircraft accidents .

Step-by-step explanation:

Given : Mean number of aircraft accidents = 9 per month

The Poisson distribution formula :-

$\dfrac{e^{-\lambda}\lambda^x}{x!}$ , where $\lambda$ is the mean of the distribution.

If X = the number of aircraft accidents per month, then the probability that in a month, there will be no aircraft accidents will be :-

$\dfrac{e^{-9}(9)^0}{0!}=0.000123409804087\approx0.0001$

Hence, the probability that in a month, there will be no aircraft accidents = 0.0001

Since this is less than 0.5 , therefor it is unlikely to have a month with no aircraft accidents .

5 0

3 years ago