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umka2103 [35]
3 years ago
7

Jayla starts with a population of 100 amoebas that increases 40% every hour for a number of hours, h. The expression 100(1 + 0.4

)h finds the number of amoebas after h hours. Which statement about this expression is true?
Mathematics
2 answers:
iren2701 [21]3 years ago
6 0

Answer:

d   It is the product of the initial population and the growth factor after h hours.

Step-by-step explanation:


Ivanshal [37]3 years ago
5 0
I don't believe your expression is correct as is; it should be <span>100(1 + 0.4)^h.

You have not shared the possible answer choices; please do so.


</span>
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A poultry farmer wishes to estimate the average incubation period (the number of days between a hen laying her egg and the time
sergiy2304 [10]

Answer:

49 eggs.

Step-by-step explanation:

The formula for Margin of Error =

z × standard deviation/√number of samples

z = z score of 98% confidence interval = 2.326

Margin of Error = Half a day = 1/2day = 0.5 day

Standard deviation = 1.5 days

Number of samples = number of eggs he needs to sample = unknown.

Imputing these above values into the formula

Margin of Error = z × standard deviation/√number of samples

0.5 = 2.326 × 1.5/√n

Cross Multiply

0.5 × √n = 2.326 × 1.5

√n = 2.326 × 1.5/0.5

√n = 3.489/0.5

√n = 6.978

Square both sides

(√n)² = 6.978²

n = 48.692484

n ≈ Approximately to the nearest whole number = 49

Therefore, the number of eggs he needs to sample to create the desired interval is approximately to the nearest whole number 49 eggs

8 0
3 years ago
Two researchers conducted a study in which two groups of students were asked to answer 42 trivia questions from a board game. Th
denpristay [2]

Answer:

(1) Correct option is B.

(2) Correct option is C.

Step-by-step explanation:

The information provided is:

n_{1}=200,\ \bar x_{1}=22.7,\ s_{1} = 4.5\\n_{2}=200,\ \bar x_{2}=19.7,\ s_{2} = 4.3

The (1 - <em>α</em>)% confidence interval for the difference between two mean is:

CI=\bar x_{1}-\bar x_{2}\pm t_{\alpha/2, (n_{1}+n_[2}-2}\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}} }

The critical value of <em>t</em> is:

\alpha /2=0.05/2=0.025

degrees of freedom =n_{1}+n_{2}-2=200+200-2=398

t_{\alpha/2, n_{1}+n_{2}-2}=t_{0.025, 398}=1.96

Compute the 95% confidence interval for the difference between two mean as follows:

CI=22.7-19.7\pm 1.96\sqrt{\frac{4.5^{2}}{200}+\frac{4.3^{2}}{200} }\\=3\pm0.8624\\=(2.1376, 3.8624)\\\approx(2.14, 3.86)

Thus, the 95% confidence interval, (2.14, 3.86) implies that the true mean difference value is contained in this interval with probability 0.95.

Correct option is B.

The null value of the difference between means is 0.

As the value 0 is not in the interval this implies that there is a difference between the two means, concluding that priming does have an effect on scores.

Correct option is C.

4 0
3 years ago
Write the equation of line in slope-intercept form. Line parallel to y=0.5x+3 that passes through the point (−9,12)
Sliva [168]

A line parallel to the given one will have the same slope, 0.5. For the purpose here, it is convenient to start with a point-slope form of the equation, then simplify. For slope m and point (h, k), the equation of the line can be written as

... y = m(x -h) +k

We have m=0.5, (h, k) = (-9, 12), so the equation is ...

... y = 0.5(x +9) +12

... y = 0.5x +16.5

8 0
3 years ago
Read 2 more answers
0+oi, 4+0iand 0+3i the area and the length
Nuetrik [128]
What's the question you're asking?
5 0
3 years ago
Solve 2.02W = -3.636
ValentinkaMS [17]

Answer: W=-1.8

Step-by-step explanation:

Divide both sides by 2.02 and you are left with W=-1.8.

Hope this helps!

7 0
3 years ago
Read 2 more answers
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