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denis-greek [22]
3 years ago
5

PLSSSSS HELPP MEEEEEEEEE PLS RESPOND TO ME

Mathematics
1 answer:
aivan3 [116]3 years ago
8 0

Answer:

HOLAAAAAO

Step-by-step explanation:

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PLEASE HELP ASAP FOR TEST
Sauron [17]
i believe the answer is C.) 0.8840
if im wrong im sorry
5 0
3 years ago
Whats an outlier?
ANTONII [103]

Answer:

In statistics, an outlier is a data point that differs significantly from other observations.

Step-by-step explanation:

An outlier may be due to variability in the measurement or it may indicate experimental error; the latter are sometimes excluded from the data set. An outlier can cause serious problems in statistical analyses.

3 0
3 years ago
A triangle has side lengths of (10x+7y)(10x+7y) centimeters, (3x+3z)(3x+3z) centimeters, and (8z+10y)(8z+10y) centimeters. Which
lora16 [44]

Answer:

(109x²+140xy+18xz+49y²+160yz+73z²) centimeters

Step-by-step explanation:

perimeter = sum of all sides

side A = (10x+7y)(10x+7y)

= 100x² +70xy +70xy +49y²

= (100x² +140xy +49y²) centimeters

side B = (3x+3z)(3x+3z)

= 9x² +9xz +9xz +9z²

= (9x² +18xz +9z²) centimeters

side C = (8z+10y)(8z+10y)

= 64z² +80yz +80yz +100y²

= (64z² +160yz +100y²) centimeters

Perimeter = 100x²+140xy+49y² + 9x²+18xz+9z² + 64z²+160yz+100y²

collect like terms

= 100x² +9x²+140xy+18xz+49y²+160yz+9z²+64z²

= (109x²+140xy+18xz+49y²+160yz+73z²) centimeters

7 0
3 years ago
Read 2 more answers
Marcy had 9.98 pounds of sauce. She wants to put them into 2 equal bags. How many pounds of sauce will be in one bag?
aksik [14]
If Marcy would want to have 2 bags with the equal amount of sauce from a 9.98 pound bag of sauce, then all she needs to do is to divide the total amount of sauce, which is 9.98, by 2 bags. So if she would do this, then the equation will be 9.98/2. The answer to this problem would be 4.99 pounds of sauce per bag. This makes since because 4.99 pounds (in the first bag) + 4.99 pounds (in the second bag) will make up 9.98 pounds which is the total amount of sauce Marcy has. 
5 0
3 years ago
Read 2 more answers
You flip a coin that is not fair, the prbability of heads on each flip is 0.7. if the coin shows heads, you draw a marble from u
Shkiper50 [21]

Solution :

P(H) = 0.7  ;  P(T) = 0.3

If heads, then Urn H,  1 blue and 4 red marbles.

If tails, then Urn T ,  3 blue and 1 red marbles.

a).

P ( choosing a Red marble )

= P (H) x P( Red from Urn H) + P (T) x P( Red from Urn T)

$=0.7 \times  \frac{4}{5} + 0.3 \times \frac{1}{4}$

= 0.56 + 0.075

= 0.635

b).  If P (B, if coin showed heads)

If heads, then marble is picked from Urn H.

Therefore,

P (Blue) $=\frac{1}{5}$

             = 0.2

c). P (Tails, if marble was red)

$=P (T/R) = \frac{P(R/T)}{P(R)}  \ P(T)$

Where P (R/T) = P ( red, if coin showed tails)

                        $=\frac{1}{4}$

                        = 0.25 (As Urn T is chosen)

P (R) =  P (Red) = 0.635 (from part (a) )

P (T) = P (Tails) = 0.3

∴ $P(T/R) = \frac{0.25 \times 0.3}{0.635}$

                = 0.118

                 

8 0
3 years ago
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