3/4 of a piece of metal has a mass of 15kg.what is the mass of 2/5 of the piece of metal

The number of bacteria in a petri dish on the first day was 113 cells. If the number of bacteria increase at a rate of 82% per d

Tresset [83]

Answer:

4107 cells

Step-by-step explanation:

From the question, we have the following values:

Day 1 : 113 cells

Number of cells increases by day by 82%

Hence,

Day 2

113 × 82% = 92.66cells

Hence, Total number of bacteria cells for Day 2 = 113 + 92.66 = 205.66cells

Day 3

205.66 × 82% = 168.6412 cells

Hence, Total number of bacteria cells for Day 3 = 168.6412 + 205.66 = 374.3012 cells

Day 4

374.3012 × 82% = 306.926984 cells

Hence, Total number of bacteria cells for Day 4 = 306.926984 + 374.3012 = 681.228184 cells

Day 5

681.228184 × 82% = 558.60711088 cells

Hence, Total number of bacteria cells for Day 5 = 558.60711088 + 681.228184 = 1239.8352949 cells

Day 6

1239.8352949 × 82% = 1016.6649418 cells

Hence, Total number of bacteria cells for Day 5 = 1016.6649418 + 1239.8352949 = 2256.5002367 cells

Day 7

2256.5002367 × 82% = 1850.3301941 cells

Hence, Total number of bacteria cells for Day 7 = 1850.3301941 + 2256.5002367 = 4106.8304308 cells

Approximately to nearest whole number, the total number of bacteria cells that would be present after 7 days = 4107 cells

3 0

2 years ago

The pucks used by the National Hockey League for ice hockey must weigh between and ounces. Suppose the weights of pucks produced

Dahasolnce [82]

Answer:

$P(5.5$

And we can find this probability using the normal standard distribution or excel and we got:

$P(-2.769$

Step-by-step explanation:

For this case we assume the following complete question: "The pucks used by the National Hockey League for ice hockey must weigh between 5.5 and 6 ounces. Suppose the weights of pucks produced at a factory are normally distributed with a mean of 5.86 ounces and a standard deviation of 0.13ounces. What percentage of the pucks produced at this factory cannot be used by the National Hockey League? Round your answer to two decimal places. "

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(5.86,0.13)$

Where $\mu=5.86$ and $\sigma=0.13$

We are interested on this probability

$P(5.5$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(5.5$

And we can find this probability using the normal standard distribution or excel and we got:

$P(-2.769$

4 0

2 years ago

The volume V (in cubic feet) of a right cylinder with a height of 3 feet and radius r (in feet) is given by V=3πr2. Solve the fo

makkiz [27]

Answer:

118. The radius is 118 feet.

Step-by-step explanation:

6 0

2 years ago

Factor 16x2 + 121 over the set of complex numbers.<br> Enter your answer in the box.

aliya0001 [1]

Answer:

(4x - 11i)(4x -+11i)

Step-by-step explanation:

Factor as a difference of squares

a² - b² = (a - b)(a + b)

note that i² = - 1

Given

16x² + 121

= (4x)² - (11i)²

= (4x - 11i)(4x + 11i)

5 0

2 years ago

N(17+x)=34x−r<br> I need to solve for x

Paraphin [41]

Answer:

The value of x is, $x= \frac{17N+r}{34-N}$

Explanation:

Given: $N(17+x)=34x-r$

Distributive Property states that when a number is multiplied by the sum of two numbers, the first number can be distributed to both of those numbers and multiplied by each of them separately.

If $a\cdot(b+c) =a\cdot b + a\cdot c$