54 ?

Suppose that the bacteria in a colony grow unchecked according to the Law of Exponential Change. The colony starts with 1 bacter

inn [45]

Answer: There are 2.25×10³⁴ bacteria at the end of 24 hours.

Step-by-step explanation:

Since we have given that

Number of bacteria initially = 1

It triples in number every 20 minutes.

So, $\dfrac{20}{60}=\dfrac{1}{3}$

So, our equation becomes

$y=y_0e^{\frac{1}{3}k}\\\\3=1e^{\frac{1}{3}k}\\\\\ln 3=\dfrac{1}{3}k\\\\k=\dfrac{1.099}{0.333}=3.3$

We need to find the number of bacteria that it will contain at the end of 24 hours.

So, it becomes,

$y=1e^{24\times 3.3}\\\\y=e^{79.1}\\\\y=2.25\times 10^{34}$

Hence, there are 2.25×10³⁴ bacteria at the end of 24 hours.

3 0

3 years ago

What would be the measure of angle DGF, if segment DG is a bisector of angle G

olga2289 [7]

Check the picture below.

a bisector, means, it "cuts in two equal halves".

6 0

3 years ago

Ned,Tony,Matt and juan are palyung basketball. Ned scores 2p+3 points more than Ned, Matt scored teice as many points as Tony, a

horsena [70]

<span>Ned scored 2p+3. Tony scored 3 points more than Ned, so Tony scored 3+2p+3, or 2p+6. Matt scored twice as many points as Tony, so Matt scored 2(2p+6), or 4p+12. Juan scored 8 fewer points than Ned, so Juan scored 2p+3-8, or 2p-5. To simplify, we add these four values. 2p + 3 2p + 6 4p + 12 2p - 5 This equals 10p + 16.</span>

7 0

3 years ago

The USDA conducted tests for salmonella in produce grown in California. In an independent sample of 252 cultures obtained from w

Marat540 [252]

Answer:

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There no sufficient evidence to show that the proportion of salmonella in the region’s water differs from the proportion of salmonella in the region’s wildlife

Step-by-step explanation:

From the question we are told that

The first sample size is $n_1 = 252$

The number that tested positive is $k_1 = 18$

The second sample size is $n_2 = 476$

The number that tested positive is $k_2 = 20$

The level of significance is $\alpha = 0.01$

Generally the first sample proportion is mathematically represented as

$\^ p _1 = \frac{k_1 }{ n_1 }$

=> $\^ p _1 = \frac{18 }{ 252 }$

=> $\^ p _1 = 0.071$

Generally the second sample proportion is mathematically represented as

$\^ p _2 = \frac{k_2 }{ n_2 }$

=> $\^ p _2 = \frac{20 }{ 476}$

=> $\^ p _2 = 0.042$

The null hypothesis is $H_o : p_1 - p_2 = 0$

The alternative hypothesis is $H_a : p_1 - p_2 \ne 0$

Generally the test statistics is mathematically represented

$z = \frac{ \^ p_1 - \^ p_2 - ( p_ 1 - p_2 )}{ \sqrt{\frac{\^ p_1 (1-\^ p_1)}{ n_1 } + \frac{\^ p_2 (1-\^ p_2)}{ n_2 } } }$

=> $z = \frac{ 0.071 - 0.042 - 0 }{ \sqrt{\frac{0.071 (1-0.071)}{ 252 } + \frac{0.042 (1-\^ 0.042)}{ 476 } } }$

=> $z = 1.56$

From the z table the area under the normal curve to the right corresponding to 1.56 is

$P(Z > 1.56 ) =0.05938$

Generally the p-value is mathematically represented as

$p-value = 2 * P(Z > 1.56 )$

=> $p-value = 2 * 0.05938$

=> $p-value = 0.1188$

From the value obtained we see that $p-value > \alpha$ hence

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There no sufficient evidence to show that the proportion of salmonella in the region’s water differs from the proportion of salmonella in the region’s wildlife

5 0

3 years ago

Write the equation of the circle with center (2, 1) and (5, 2) a point on the circle.

Y_Kistochka [10]

(x-2)^2 +(y-1)^2 = 10

6 0

4 years ago