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Sergio [31]
2 years ago
8

Janet can read 3.6 pages of a book in a minute. If she read for 2.71 minutes. How much would have she had read

Mathematics
1 answer:
Aliun [14]2 years ago
8 0

Answer:

9.76 pages

Step-by-step explanation:

multiply 3.6 times 2 to get 7.2. Then multiply 3.6 times .71 to get 2.56 then add 7.2 and 2.56 to get <u>9.76</u> as your answer.

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Find the value of z such that 0.76980.7698 of the area lies between −z−z and z. round your answer to two decimal places.
babunello [35]
P(-z<Z<z)=0.7698
P(0<Z<z)=0.7698/2=0.3849
P(Z<0)=0.5
P(Z<z)=P(Z<0)+P(0<Z<z)=0.5+0.3849=0.8849
Using the z-table, the z-score that corresponds to probability of 0.8849 is z=1.20 
4 0
2 years ago
Read 2 more answers
PLEASE HELP ASAP Eva is at a sushi restaurant. She ordered 2 pieces of squid for a total of $3.50, 1 piece of eel for $3.25, 3 p
mrs_skeptik [129]

Answer:

I do not have enough time to put them in order if you don't mind but i can list the prices.

Step-by-step explanation:

each piece of squid is 1.25

each piece of tuna is 3.25

each piece of crab is 2.00

hope this helps <3

4 0
2 years ago
What is 10.2719 rounded to the nearest hundreth?
Paladinen [302]
Find the number in the hundredth place
7
7
and look one place to the right for the rounding digit
1
1
. Round up if this number is greater than or equal to
5
5
and round down if it is less than
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10.27
4 0
2 years ago
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A teacher bought 7 packs of dominoes for a math game. Each pack had 55 dominoes. The teacher already had 118 dominoes to use in
S_A_V [24]

Answer:

540540 dominoes

Step-by-step explanation:

6 0
2 years ago
The functions $f$ and $g$ are defined as follows: \[f(x) = \sqrt{\dfrac{x+1}{x-1}}\quad\text{and}\quad g(x) = \dfrac{\sqrt{x+1}}
CaHeK987 [17]

Answer:

The answer is "domain and range are different".

Step-by-step explanation:

Given:

f(x) = \sqrt{\frac{x+1}{x-1}}\\\\g(x) =\frac{\sqrt{x+1}}{\sqrt{x-1}}\\

Solve f(x) to find domain and range:

for element x: R:

⇒ x \leq -1 \ \ \ and \  \ \  x > 1 range:

for  \ \ f(x) \times element : 0 \leq f(x)< 1 \ \ \ or \ \ f(x)>1

g(x) =\frac{\sqrt{x+1}}{\sqrt{x-1}}\\

Solve for domain:

⇒ \ when \ x \ element \ R: \ \ \ x>1  

Solve for range:  

⇒ g(x) \times element  R : g(x)>1

So, the value of the method f(x) and g(x) (range and domain) were different.  

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2 years ago
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