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Andreyy89
3 years ago
15

2/3 (5 x 4) ÷ 3 + 6/7 =

Mathematics
2 answers:
Kobotan [32]3 years ago
8 0
Its 5.30158730159 on a calculator.. So if you round it most likely.. Im guessing itll be 5.30 ?.. Im not so sure.. 
Lerok [7]3 years ago
6 0
334/63
if u want the mixed fraction it'd be 5(19/63)
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Margie is practicing for an upcoming tennis tournament. Her first serve is good 20 out of 30 times on average. Margie wants to k
mojhsa [17]

Answer:

0.6804

Step-by-step explanation:

Given that Margie is practicing for an upcoming tennis tournament. Her first serve is good 20 out of 30 times on average.

Since each trial is independent and there are two outcomes, X no of good serves is binomial with n=6 and p =2/3

Required probability

= Prob that atleast four of 6 times good serve

=P(X\geq 4)

=P(x=4,5 or 6)\\=1-F(3)\\=0.6804

Formula used:

P(X=r) =6Cr (\frac{2}{3}^r )(\frac{1}{3}^{6-r}  )

7 0
3 years ago
Read 2 more answers
60 is whatpercent of 40
DiKsa [7]

<u>Answer:</u>

24

<u>Step-by-step explanation:</u>

Y is 60% of 40

Equation: Y = P% * X

<u>Solving our equation for Y</u>

Y = P% * X

Y = 60% * 40

<u>Converting percent to decimal:</u>

p = 60%/100 = 0.6

Y = 0.6 * 40

Y = 24

8 0
3 years ago
A mathematical sentence which contains an inequality symbol and one variable raised to the first power is called a ____ inequali
ch4aika [34]
The answer to fill in the blank should be ''Linear''.
5 0
3 years ago
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Mkey [24]
The answer to this question is the second choice, jessica bought 1/4 more than derek

6 0
2 years ago
For the function f(x)=(1/2)⋅6^x , identify the domain, range, y-intercept, and asymptote.
skelet666 [1.2K]

Answer:

Step-by-step explanation:

f(x) = (1/2) *6^x = 2^-1 * 2^x *3^x

y= 2^ (x-1) * 3^x

Domain represents all values that x can have and is all real numbers

Range all values that y can have and is y > 0 ( all real numbers that are positive)

The y-intercept is the point where the graph intersects the y-axis so x=0 there

y = 2^(0-1) *3^0 = 1/2 *1 = 1/2

the asymptote is at y= 0

6 0
3 years ago
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