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hodyreva [135]
3 years ago
11

How many hundredths are in 539.26? Is the answer 53,926

Mathematics
2 answers:
Nezavi [6.7K]3 years ago
8 0

Answer:

53926

Step-by-step explanation:

539.26x100=53926

Angelina_Jolie [31]3 years ago
6 0

Answer:

there are 6 hundredths

Step-by-step explanation:

the 6 is in the hundredths decimal place so the answer would be 6

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Briainliest Simplify 15√90/√125<br> options<br> 9√2<br> 18√5<br> 18√2/5<br> 27√2/5
nekit [7.7K]

Answer:

9√2

Step-by-step explanation:

simplify the radical by breaking the radicand up into a product of known factors, assuming positive real numbers.

Exact form:9√2

Decimal form: 12.72792206

3 0
3 years ago
If g(x)=7x²−4x−3 and h(x)=5x²−12x, find (h∘g)(1)<br><br><br>FIRST BEST ANSWER BRAINLIEST
AURORKA [14]

Answer: See below

Step-by-step explanation:

h\:\circ \:g=h\left(g\left(x\right)\right)

=5\left(7x^2-4x-3\right)^2-12\left(7x^2-4x-3\right)

=5\left(7x^2-4x-3\right)\left(7x^2-4x-3\right)-12\left(7x^2-4x-3\right)

=245x^4-280x^3-130x^2+120x+45-84x^2+48x+36

Simplify:

=245x^4-280x^3-214x^2+168x+81

6 0
3 years ago
If driving 60 mph for 9 hours how many miles will they travel in a day
Dimas [21]
60x9=540 I believe is the answer
8 0
3 years ago
Find the distance between (-8,4) and (-8,-2)
densk [106]

Answer:

+6

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
In microbiology, colony-forming units (CFUs) are used to measure the number of microorganisms present in a sample. To determine
zimovet [89]

Answer:

(a)  0.2650

(b)  0.0111

(c)  0.0105

(d)  0.0006

Step-by-step explanation:

Given that:

In microbiology, colony-forming units (CFUs) are used to measure the number of microorganisms present in a sample.

Suppose that the number of CFUs that appear after incubation follows a Poisson distribution with mean μ = 15.       &;

If the area of the agar plate is 75cm²;

what is the probability  of observing fewer than 4 CFUs in a 25 cm² area of the plate.

We can determine the mean number of CFUs that appear on a 25cm² area of the plate as follows;

75cm²/25cm² = 3

Since;

mean  μ = 15  

mean number of CFUs that appear on a 25cm² = 15/3 = 5 CFUs

Thus ; the probability of observing fewer than 4 CFUs in a 25 cm² area of the plate is estimated as:

= P(X < 4)

Using the EXCEL FUNCTION ( = poisson.dist(3, 5, TRUE) )

we have ;

P(X < 4) = 0.2650

b) If you were to count the total number of CFUs in 5 plates, what is the probability you would observe more than 95 CFUs?

Given that the total number of CFUs = 5 plates; then the mean number of CFUs in 5 plates =  15×5 = 75 CFUs

The probability is therefore = P( X > 95 )

= 1 - P(X ≤ 95)

= 1 - poisson.dist(95,75,TRUE) ( by using the excel function)

= 0.0111

c) Repeat the probability calculation in part (b) but now use the normal approximation.

Let assume that the mean and the variance of the poisson distribution are equal

Then;

X \sim N (\mu = 75 , \sigma^2 = 75)

We are to repeated the probability calculation in part (b) from above;

So:

P( X > 95 )

use the normal approximation

From standard normal variable table:

P(Z > 2.3094)

Using normal table

P(Z > 2.3094) = 0.0105

(d)  Find the difference between this value and your answer in part (b).

So;

the difference between the value in part c and part b is;

=  0.0111 - 0.0105

= 6*10^{-4}

= 0.0006 to four decimal places

5 0
4 years ago
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