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3 years ago

If a = 4 and b = 7, what is the value of the following expression? 20 ÷ a + 7 • b A. 28 B. 84 C. 54 D. 33

Mathematics

2 answers:

expeople1 [14]3 years ago

8 0

Answer:

Step-by-step explanation:

20 ÷ a + 7 • b

Let a = 4 and b = 7

20 ÷ 4 + 7 • 7

Multiply and divide first

5 + 49

Then add

ladessa [460]3 years ago

4 0

Answer:

Step-by-step explanation:

a=4 b=7

20/4=5

7✖️7=49

5+49=54

Hope this helps you!

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The population of a colony of mosquitoes obeys the law of uninhibited growth. If there are 1000 mosquitoes initially and there a

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Answer:

1) The size of the colony after 4 days is 6553 mosquitoes

2) After $t = 4.9\ days$

Step-by-step explanation:

To answer this question you must use the growth formula

$N = N_0e ^ {kt}$

Where

$N_0$ is the initial population of mosquitoes = 1000

t is the time in days

k is the growth rate

N is the population according to the number of days

We know that when t = 1 and $N_0=1000$ then N = 1600

Then we use these values to find k.

$1600 = 1000e ^{k(1)}\\\\\frac{1600}{1000} = e ^ k\\\\ln(\frac{1600}{1000}) = k\\\\k = 0.47$

Now that we know k we can find the size of the colony after 4 days.

$N = 1000e ^ {0.47(4)}\\\\N = 6553\ mosquitoes$

To know how long it should take for the population to reach 10,000 mosquitoes we must do N = 10000 and solve for t.

$10000 = 1000e ^ {0.47t}\\\\10 = e ^ {0.47t}\\\\ln(10) = 0.47t\\\\t = \frac{ln(10)}{0.47}\\\\t = 4.9\ days$

6 0

2 years ago

A bus driver covers a certain distance in 3 hours at an average speed of 80 km/h. How long will the journey take at an average s

BaLLatris [955]

Answer:

d= s x t

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2 years ago

A study sampled 350 upperclassmen (Group 1) and 250 underclassmen (Group 2) at high schools around the city of Portland. The stu

iVinArrow [24]

Answer:

$-0.061 < P_1 -P_2< 0.025$

Step-by-step explanation:

Give data:

$n_1 = 350$

$n_2 =250$

$x_1 = 23$

$x_2 = 21$

$\hat{P} 1 = \frac{x_1}{n_1} = \frac{23}{350} = 0.066$

$\hat{P} 2 = \frac{x_2}{n_2} = \frac{21}{250} = 0.084$

for 95% confidence interval

$\alpha = 1 - 0.95 = 0.05 and \alpha/2 = 0.025$

$z_{\alpha/2} = 1.96$ from standard z- table

confidence interval for P_1 and P_2 is

$\hat{P} 1 - \hat{P} 2 \pm z_{\alpja/2} \sqrt{\frac{\hat{P} 1(1-\hat{P} 1)}{n_1} +\frac{\hat{P} 2(1-\hat{P} 2)}{n_2}}$

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$-0.018 \pm 0.043$

confidence interval is

$-0.018 - 0.043 < P_1 -P_2$

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Step-by-step explanation:

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