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professor190 [17]
3 years ago
13

What's the answer for this?

Mathematics
1 answer:
il63 [147K]3 years ago
6 0
X-(-7)=48 I believe this correct
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89 squared. The hypotenuse is equal to 9.434 which is 89 squared.
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Can anyone help me with this question
pashok25 [27]

Log base 3 of 4 + log base 3 of 11= log base 3 of 4•11; Answers is 11

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I need help ASAP yall
BARSIC [14]

Answer:

i would say C

Hope This Helps!   Have A Nice Night!!

7 0
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Read 2 more answers
The Central Limit Theorem tells us that for population distribution(s), if we repeatedly take new random samples from this distr
sertanlavr [38]

Answer:

d. If your sample size is very large, the distribution of the sample averages will look more like distribution.

Step-by-step explanation:

The central limit Theorem states that for population distribution if you repeatedly take samples from the distribution, then the normal thing for it to happen would be that the distribution means of the samples will be normally distributed, this is what it states, the option that comes closer to that statement would be  d. If your sample size is very large, the distribution of the sample averages will look more like distribution, because they large sample will create for a normally distributed means distribution.

7 0
3 years ago
) Using the chart, approximate the limit of the function f(x) = sin x/ x as x approaches zero. Note the numbers are in radians:
Rashid [163]

Answer:

1

Step-by-step explanation:

We are given that

f(x)=\frac{sinx}{x}

Numbers are in radians

Substitute x=-1

f(-1)=\frac{sin(-1)}{-1}=0.84

Substitute x=-0.25

f(-0.25)=\frac{Sin(-0.25)}{-0.25}=0.989

Substitute x=-0.01

f(-0.01)=\frac{sin(-0.01)}{-0.01}=0.999

Substitute x=-0.005

f(-0.005)=\frac{sin(-0.005)}{-0.005}=0.999

Substitute x=0.005

f(0.005)=\frac{sin(0.005)}{0.005}=0.999

Substitute x=0.01

f(0.01)=\frac{sin(0.01)}{0.01}=0.999

Substitute x=0.25

f(0.25)=\frac{sin(0.25)}{0.25}=0.989

Substitute x=1

f(1)=\frac{sin(1)}{1}=0.84

Therefore, \lim_{x\rightarrow 0}\frac{sinx}{x}=1

8 0
3 years ago
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