Ask question

Ask question

All categories

2 years ago

7

each hour the number of bacteria in dr. nall's petri dish tripled what percent is the population compared to the population the

hour before

1 answer:

yan [13]2 years ago

3 0

The starting percent is 100%. After one hour, bacteria tripled, so it is 300%.

You might be interested in

Solve for t <br><br> t/5/12=-3/10

natka813 [3]

T = -18 hope this helped

6 0

2 years ago

in the diagram,the pyramid has a rectangular base.ifthe pyramid has a volume of 176 cmcube and the height of 11cm ,find the valu

SOVA2 [1]

The formula of pyramid:
$v = \frac{1}{3} ah$
Where v is the volume, a is the base area and h is the height.

We can then put the given information into the formula:

176 = (1/3)(3x × x)(11)

(176 × 3) / 11 = 3x^2

3x^2 = 48

x^2 = 48/3

x = _/16

x = 4

3 0

3 years ago

There are 12 donuts in a box, and Claire at 2. Approximately what percent of the donuts remain in the box? (Round to the nearest

skelet666 [1.2K]

Answer:

83% remains in the box

Step-by-step explanation:

Here, we are interested in calculating the percentage of donuts remaining in the box.

From the question, we are told that 2 was eaten, the number of donuts left = 12-2 = 10 donuts

The percentage of donuts left would be;

Number of donuts left/Total no of donuts * 100%

= 10/12 * 100 = 83.33

Which is 83% to the nearest whole number percentage

5 0

3 years ago

Write the percent as a decimal.<br> 0.06%<br> PLS HELPPPPPP

Studentka2010 [4]

Answer:

0.0006

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

A single die is rolled. Find the probability of rolling an oddnumber or a number less than 6.

mina [271]

Given

A single die is rolled.

To find the probability of rolling an odd number or a number less than 6.

Explanation:

It is given that,

A single die is rolled.

Then, the sample space is,

$\begin{gathered} S=\lbrace1,2,3,4,5,6\rbrace \\ n(S)=6 \end{gathered}$

Let A be the event of getting an odd number.

Then,

$\begin{gathered} A=\lbrace1,3,5\rbrace \\ n(A)=3 \end{gathered}$

Let B be the event of getting a number less than 6.

Then,

$\begin{gathered} B=\lbrace1,2,3,4,5\rbrace \\ n(B)=5 \end{gathered}$

Also,

$\begin{gathered} A\cap B=\lbrace1,3,5\rbrace \\ n(A\cap B)=3 \end{gathered}$

Therefore,

$\begin{gathered} P(A\cup B)=P(A)+P(B)-P(A\cap B) \\ =\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{n(A\cap B)}{n(S)} \\ =\frac{3}{6}+\frac{5}{6}-\frac{3}{6} \\ =\frac{3+5-3}{6} \\ =\frac{5}{6} \end{gathered}$

Hence, the answer is 5/6.

5 0

11 months ago

Read 2 more answers