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uysha [10]
3 years ago
10

Square root 5 is closest to what real number?

Mathematics
1 answer:
gladu [14]3 years ago
7 0
The square root of 5 is 2.23 so it would be closer to 2
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Incomes in a certain town are strongly right-skewed with mean $36,000 and standard deviation $7000. a random sample of 75 househ
Svet_ta [14]
Since the sample is greater than 10, we can approximate this binomial problem with a normal distribution.

First, calculate the z-score:

z = (x - μ) / σ = (37000 - 36000) / 7000 = 0.143

The probability P(x > 37000$) = 1 - P(<span>x < 37000$), 
therefore we need to look up at a normal distribution table in order to find 
P(z < 0.143) = 0.55567 
And 
</span>P(x > 37000$) = 1 - <span>0.55567 = 0.44433

Hence, there is a 44.4% probability that </span><span>the sample mean is greater than $37,000.</span>

4 0
3 years ago
if a bathtub is being filled at a rate of 2.5 gallons per minute, how long will it take to fill the bathtub?
umka2103 [35]

You need say how big the bathtub it is or you cant do it.

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3 0
3 years ago
Read 2 more answers
PLEASE PLEASE HELP ME! thank you so so much
viva [34]

Answer:

6π

Step-by-step explanation:

(135/360)×π×4²

(135/360)×π×16

=6π

6 0
3 years ago
The mode of the data above is 6
nordsb [41]
I was stuck on the same thing in my class test. I ended up failing but if I get the answers to it I’ll totally send them to you
7 0
3 years ago
]solomon needs to justify the formula for the arc length of a sector. which expression best completes this argument? the circumf
Anna35 [415]

Answer:

\frac{2 \pi r}{\frac{360^{\circ}}{n^{\circ}}} best completes this argument

Step-by-step explanation:

Circumference of circle =\pi \cdot d

Where d is the diameter of circle

We are given that if equally sized central angles, each with a measure of n°, are drawn, the number of sectors that are formed will be equal to \frac{360^{\circ}}{n^{\circ}}

So, Number of sectors =  \frac{360^{\circ}}{n^{\circ}}

The arc length of each sector is the circumference divided by the number of sectors

\Rightarrow \frac{\pi \cdot d}{\frac{360^{\circ}}{n^{\circ}}}

Diameter d = 2r (r = radius)

\Rightarrow \frac{2 \pi r}{\frac{360^{\circ}}{n^{\circ}}}

Option b is true

Hence\frac{2 \pi r}{\frac{360^{\circ}}{n^{\circ}}} best completes this argument

7 0
3 years ago
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