All categories

3 years ago

DUE TOMORROW PLEASE HELP!!!!

2 answers:

7 0

1 day = 24 hours
1 day = 103 680
therefore 24 hours = 103 680
1 hour = 103 680 ÷ 24
= 4320

1 hour = 60 mins
1 hour = 4320
therefore 60 mins = 4320
1 min = 72 beats

algol133 years ago

6 0

I agree with them . Those are the ways you could do it

You might be interested in

(Edmentum)

Kisachek [45]

Answer:

kjbhgjc fytjgcyvcjyhkvchgvjhgvhjvbj

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

HELPPP WILL MARK BRAINLIEST

leva [86]

Answer:

Step-by-step explanation:

y - z+6

7 0

3 years ago

Read 2 more answers

HELP ME ASAP YOU GET 10 POINTS

snow_lady [41]

Answer:

domain 12

range 24

brainliest?

7 0

3 years ago

Read 2 more answers

Suppose 241 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea. Use a 0.05 significance

lana66690 [7]

Answer:

Null Hypothesis, $H_0$ : p = 0.20

Alternate Hypothesis, $H_a$ : p > 0.20

Step-by-step explanation:

We are given that 241 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea.

We have to use a 0.05 significance level to test the claim that more than 20% of users develop nausea.

Let p = population proportion of users who develop nausea

So, Null Hypothesis, $H_0$ : p = 0.20

Alternate Hypothesis, $H_a$ : p > 0.20

Here, null hypothesis states that 20% of users develop nausea.

And alternate hypothesis states that more than 20% of users develop nausea.

The test statistics that would be used here is One-sample z proportion test statistics.

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = proportion of users who develop nausea in a sample of 241 subjects = $\frac{54}{241}$

n = sample of subjects = 241

So, the above hypothesis would be appropriate to conduct the test.

6 0

3 years ago

How do you do this question?

Alex Ar [27]

Step-by-step explanation:

(a) dP/dt = kP (1 − P/L)

L is the carrying capacity (20 billion = 20,000 million).

Since P₀ is small compared to L, we can approximate the initial rate as:

(dP/dt)₀ ≈ kP₀

Using the maximum birth rate and death rate, the initial growth rate is 40 mil/year − 20 mil/year = 20 mil/year.

20 = k (6,100)

k = 1/305

dP/dt = 1/305 P (1 − (P/20,000))

(b) P(t) = 20,000 / (1 + Ce^(-t/305))

6,100 = 20,000 / (1 + C)

C = 2.279

P(t) = 20,000 / (1 + 2.279e^(-t/305))

P(10) = 20,000 / (1 + 2.279e^(-10/305))

P(10) = 6240 million

P(10) = 6.24 billion

This is less than the actual population of 6.9 billion.

P(100) = 7570 million = 7.57 billion

P(600) = 20,000 / (1 + 2.279e^(-600/305))

P(600) = 15170 million = 15.17 billion

7 0

3 years ago