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3 years ago

1. Pedro is a good student. a. b.

Mathematics

1 answer:

ICE Princess25 [194]3 years ago

5 0

Answer:

your answer is (A)

Step-by-step explanation:

in the affirmative

plz mark me as brainlist

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A freshly inoculated bacterial culture of Streptococcus contains 100 cells. When the culture is checked 60 minutes later, it is

Luba_88 [7]

Answer:

$P(t) = 100e^{0.0251t}$

The doubling time is of 27.65 minutes.

Step-by-step explanation:

Exponential equation of growth:

The exponential equation for population growth is given by:

$P(t) = P(0)e^{kt}$

In which P(0) is the initial value and k is the growth rate.

A freshly inoculated bacterial culture of Streptococcus contains 100 cells.

This means that $P(0) = 100$ . So

$P(t) = 100e^{kt}$

When the culture is checked 60 minutes later, it is determined that there are 450 cells present.

This means that $P(60) = 450$ , and we use this to find k. So

$450 = 100e^{60k}$

$e^{60k} = 4.5$

$\ln{e^{60k}} = \ln{4.5}$

$60k = \ln{4.5}$

$k = \frac{\ln{4.5}}{60}$

$k = 0.0251$

$P(t) = 100e^{0.0251t}$

Doubling time:

This is t for which P(t) = 2P(0) = 200. So

$200 = 100e^{0.0251t}$

$e^{0.0251t} = 2$

$\ln{e^{0.0251t}} = \ln{2}$

$0.0251t = \ln{2}$

$t = \frac{\ln{2}}{0.0251}$

$t = 27.65$

The doubling time is of 27.65 minutes.

7 0

3 years ago

In the diagram, the length of segment QV is 15 units.

Andrews [41]

Answer:

The answer is 15 units.

Step-by-step explanation:

The reason it is 15 units is because We know that QV = 15 and TQ = TS, QV = SV;4 x - 1 = 154 x = 15 + 14 x = 16x = 16 : 4 = 4 henceforth the answer being 15 units

6 0

3 years ago

Read 2 more answers

last month Tara work 16.5 hours the first week 19 hours the second week 23 hours the third week 15.75 hours the fourth week she

STatiana [176]

The number of hours, h, Tara plans to work this month are greater than 16.5 + 19 + 23 + 15.75. Therefore, h > 16.5 + 19 + 23 + 15.75.

h > 74.25

7 0

3 years ago

How many inches is 35 yards?

Tomtit [17]

Answer:

1260 inches

Step-by-step explanation:

Calculator

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2 years ago

A random sample of 16 one-kilogram sugar packets is obtained and the actual weights of the packets are measured. The sample mean

elena55 [62]

Answer:

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

Step-by-step explanation:

We are in posession of the sample's standard deviation, so we use the student's t-distribution to find the confidence interval.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 16 - 1 = 15

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 35 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.905([tex]t_{995}$ ). So we have T = 2.9467

The margin of error is:

M = T*s = 2.9467*0.058 = 0.171

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 1.053 - 0.171 = 0.882kg

The upper end of the interval is the sample mean added to M. So it is 1.053 + 0.171 = 1.224 kg.

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

3 0

3 years ago