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Lilit [14]
3 years ago
13

Help Plz!!!

Mathematics
2 answers:
il63 [147K]3 years ago
8 0
You take 300 - 99 you get 201 so you still have 201lbs so then 3 times 67 is 201 so it would be B

DedPeter [7]3 years ago
5 0

Answer :I guess that the answer is B. did this in my exaam and got it right

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Which line is it? help asap
ki77a [65]
The correct answer is line q
8 0
2 years ago
Read 2 more answers
Consider a t distribution with 7 degrees of freedom. Compute P(-1.29 < t < 1.29). Round your answer to at least three deci
marusya05 [52]

Answer:

Step-by-step explanation:

Given that there is a t distribution with 7 degrees of freedom.

P(-1.29 < t < 1.29)=1-0.2380

=0.7620

b) Now there is a different t distribution with 18 df.

When P(t\leq c)=0.05

c=-1.733

7 0
2 years ago
(04.01) Which point could be removed in order to make the relation a function? (4 points) {(0, 2), (3, 8), (−4, −2), (3, −6), (−
umka21 [38]
A function will not have any repeating x values...it can have repeating y values, just not the x ones

{(0,2),(3,8),(-4,-2),(3,-6),(-1,8),(8,3)}
u would have to remove one of the sets of points that has 3 as its x value....so either remove (3,8) or (3,-6)....because with both of them in there, u have repeating x values
7 0
3 years ago
Read 2 more answers
2. Find the area. The figure is not drawn to scale.<br> 4 cm<br> 17 cm
BabaBlast [244]
area=base x height so i’m thinking it’s 4x17=68
7 0
2 years ago
How do you solve this equation?
Travka [436]
Exponentail thingies

easy, look at all them them, see that they have 5 in common?
rremember how esay it was to factor
ax^2+bx+c=0

now we have
5^(2x)-6(5^x)+5=0
remember that 5^(2x)=(5^2)^x or (5^x)^2
in other words, we can rewrite it as
1(5^x)^2-6(5^x)+5=0
if yo want, replace 5^x with a and factor
1a^2-6a+5=0
(a-1)(a-5)=0
a=5^x
(5^x-1)(5^x-5)=0
set each to zero

5^x-1=0
5^x=1
take the log₅ of both sides
x=log₅1


5^x-4=0
5^x=4
take the log₅ of both sides
x=log₅4

x=log₅1 and/or log₅4







second quesiton

same thing

1(2^x)-10(2^x)+16=0
factor
(2^x-8)(2^x-2)=0
set each to zero

2^x-8=9
2^x=8
x=3

2^x-2=0
2^x=2
x=1

x=3 or 1







first one
x=log₅1 and/or log₅4
second one
x=1 and/or 3


7 0
3 years ago
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