Help please I need to know

Exercise 8.12 introduces data on the average monthly temperature during the month babies rst try to crawl (about 6 months after

Afina-wow [57]

Answer:

The point has a high leverage

Step-by-step explanation:

The point has a high leverage as it would act as an infinitesimal point that will have a very huge/drastic impact on the fit of the model, and this impact can be seen in such ways as listed below :

Smaller coefficient of determination
Higher sum of squares error

removing this point will eliminate these drastic impact and make the correlation to be better.

5 0

3 years ago

−2/3a+1/8−1/6a−3/4 PLEASE HELP

prohojiy [21]

Answer:

Evaluate:

- $\frac{5a}{6}$ - $\frac{5}{8}$

Factor:

$\frac{5(-4a-3}{24}$

Step-by-step explanation:

6 0

3 years ago

Given that 8 bolts and 6 nuts weigh 138grams and 3 bolts and 5 nuts weigh 71 grams. Find the weight of;(a)one bolt and one nut (

Viktor [21]

Answer:

(a) Weight of a bolt = 12 grms and weight of a nut is 7 gms. Weight of one bolt and one nut = 12 + 7 = 19 gms

(b) 69 gms

(c) 209 gms

Step-by-step explanation:

This problem relates to solution of 2 unknowns using simultaneous equations.

Let B be the weight of a single bolt and N be the weight of a single nut

Since 8 bolts and 6 nuts weigh 138 grams we get one of the equations as

8B + 6N = 138 (1)

The second equation in the unknowns is

3B + 5N = 71 (2)

To solve, we eliminate one of the unknowns by making its coefficients the same and subtracting one from the other

Multiplying (1) by 5 ===> 40B + 30N = 690 (3)

Multiplying (2) by 6 ===> 18B + 30N = 426 (4)

(3) - (4) eliminates the N variables and yields

22B = 264 ==> B = 264/22 ==> B = 12

So the weight of a single bolt is 12 grams

We can find the weight of a single nut by substituting this value of B into any of the equations (1), (2), (3) or (4) and solving for N. Let's use equation (2)

3(12) + 5N = 71 ==> 36 + 5N = 71 ==> 5N = 71-36 = 35 ==> N= 7

So the weight of a single nut is 7 grams

Weight of one bolt and one nut is the sum of the above individual weights = 12 + 7 = 19 gms

To solve (b) and (c) we could set up two other equations and plug in values for B and N

(b) 4B + 3N = 4.12 + 3.7 = 69
However, an alternate way is to perceive that 4B + 3N is exactly half of 8B + 6N so the value of that must be 138/2 = 69

(c) If we add both equations (1) and (2), we get

11B + 11N = 138 + 71 = 209 which is the equation for the total weight of 11 bolts and 11 nuts

5 0

2 years ago

GEOMETRY PROOFS!

yarga [219]

From the given figure ,

RECA is a quadrilateral

RC divides it into two parts

From the triangles , ∆REC and ∆RAC

RE = RA (Given)

angle CRE = angle CRA (Given)

RC = RC (Common side)

Therefore, ∆REC is Congruent to ∆RAC

∆REC =~ ∆RAC by SAS Property

⇛CE = CA (Congruent parts in a congruent triangles)

Hence , Proved

Additional comment:-

SAS property:-

"The two sides and included angle of one triangle are equal to the two sides and included angle then the two triangles are Congruent and this property is called SAS Property (Side -Angle-Side)

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4 0

2 years ago

Read 2 more answers

If the area of the rectangle is 42 square units, what is the area of the shaded part?

MArishka [77]

lol you also in mcp

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4 0

3 years ago