Help plzzz someone solve it i need help

Any two normal curves are the same except for their Multiple Choice standard deviations. means. standard deviations and means. s

Neko [114]

Answer:

Standard deviation, means, skewness and kurtosis.

Step-by-step explanation:

Two normal curves may be same but they have different means, standard deviation and skewness. There can be different standard deviation for two curves and there is difference in skewness.

7 0

3 years ago

Parallelogram ABCD is given. Draw line EF so that it goes through the vertex A. Point E lies on the side BC and point F lies on

tresset_1 [31]

Answer:

The proof is explained step-wise below :

Step-by-step explanation :

For better understanding of the solution see the attached figure :

Given : ABCD is a Parallelogram ⇒ AB ║ DC and AD ║ BC

Now, F lies on the extension of DC. So, AB ║ DF

To Prove : ΔABE is similar to ΔFCE

Proof :

Now, in ΔABE and ΔFCE

∠ABE = ∠FCE ( alternate angles are equal )

∠AEB = ∠FEC ( Vertically opposite angles )

So, by using AA postulate of similarity of triangles

ΔABE is similar to ΔFCE

Hence Proved.

7 0

3 years ago

gas prices decreased this week from $4 per gallon to 3.80 dollars per gallon what is the percent decrease

evablogger [386]

Answer: 5%

Is the percentage

5 0

3 years ago

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Choose the expression that represents the pattern of your sequence: 10, 20, 40, 80, ....

Angelina_Jolie [31]

It's multiplying by 2 each time 10 x 2 = 20, 20 x 2 = 40, 40 x 2 = 80.

8 0

3 years ago

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Prove that tan 10+tan 70'+taniou = tanio tanfo'tam 1oo.

Galina-37 [17]

Question:

Prove that:

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$

Answer:

Proved

Step-by-step explanation:

Given

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$

Required

Prove

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$

Subtract tan(10) from both sides

$- tan(10)+tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100) - tan(10)$

$tan(70) + tan(100) = tan(10). tan(70). tan(100) - tan(10)$

Factorize the right hand size

$tan(70) + tan(100) = -tan(10)(-tan(70). tan(100) + 1)$

Rewrite as:

$tan(70) + tan(100) = -tan(10)(1-tan(70). tan(100))$

Divide both sides by $1-tan(70). tan(100)$

$\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = \frac{-tan(10)(1-tan(70). tan(100))}{1-tan(70). tan(100))}$

$\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = -tan(10)$

In trigonometry:

$tan(A + B) = \frac{tan(A) + tan(B)}{1 - tan(A)tan(B)}$

So:

$\frac{tan(70) + tan(100)}{1 - tan(70)tan(100)}$ can be expressed as: $tan(70 + 100)$

$\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = -tan(10)$ gives

$tan(70 + 100) = -tan(10)$

$tan(170) = -tan(10)$

In trigonometry:

$tan(180 - \theta) = -tan(\theta)$

So:

$tan(180 - 10) = -tan(10)$

Because RHS = LHS

Then:

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$ has been proven

6 0

3 years ago