.5x + 2y<br> coefficients

Refer to the sequence 11, 16, 21, 26, 31, 36. Explain why the terms in the sequence will continue to alternate between even and

Debora [2.8K]

Explanation:

The common difference of this arithmetic sequence is 5. Adding 5 to an even number gives an odd number, and vice versa. Hence the "parity" of the terms in the sequence must alternate between odd and even.

8 0

3 years ago

Will give brainliest-H(t)=-16t^2+28t

____ [38]

For this case we have the following equation:
H (t) = - 16t ^ 2 + 28t
When the dolphin finishes a jump, then the height is equal to zero.
We have then:
-16t ^ 2 + 28t = 0
We look for the roots of the polynomial:
t1 = 0
t2 = 1.75 s
Therefore, the dolphin takes 1.75 seconds to make a jump.
Answer:
it will take for the dolphin to make one jump about:
t = 1.75 s

4 0

3 years ago

PLEASE PLEASE PLEASE HELP TIMEDCan you prove that DE F = HGF Justify your answer. A. Yes, the triangles are congruent by SAS. B.

LenKa [72]

Answer:

A. Yes, the triangles are congruent by SAS.

Step-by-step explanation:

EF = FG and DF = FH-> Given

angle EFD = angle HFG -> Vertical angles are congruent

DE F = HGF -> SAS Triangle Congruence Theorem

6 0

3 years ago

Read 2 more answers

If a random sample of size nequals=6 is taken from a population, what is required in order to say that the sampling distributio

goldenfox [79]

Answer:

$\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})$

In order to satisfy this distribution we need that each observation on this case comes from a normal distribution, because since the sample size is not large enough we can't apply the central limit theorem.

Step-by-step explanation:

For this case we have that the sample size is n =6

The sample man is defined as :

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And we want a normal distribution for the sample mean

$\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})$

In order to satisfy this distribution we need that each observation on this case comes from a normal distribution, because since the sample size is not large enough we can't apply the central limit theorem.

So for this case we need to satisfy the following condition:

$X_i \sim N(\mu , \sigma), i=1,2,...,n$