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Mademuasel [1]
3 years ago
6

Our hearts beat approximately 70 times per minute. Express in scientific notation how many times the heart beats over a lifetime

of 73 years​ (ignore leap​ years). Round the decimal factor in your scientific notation answer to two decimal places.
Mathematics
1 answer:
earnstyle [38]3 years ago
7 0

Answer:

2.69 x 10^{9}

Step-by-step explanation:

We are given the heart's speed - 70 bpm

We count the number of minutes in 73 years :

  • 1 year = 365 days
  • 1 day = 24 hours
  • 1 hour = 60 minutes
  • 73 x 365 x 24 x 60 = 38,368,800 minutes

We multiply the heart's bpm with 73 years worth of minutes

38,368,800 x 70 = 2,685,816,000

Write the number in scientific notation = 2.68581 x 10^{9} ≈ 2.69 x 10^{9}

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I need the answer help<br>​
lisov135 [29]

Answer:

the second one(20)

Step-by-step explanation:

If you pretend 20 is x, you solve the equation by pluging in x (20) into the equation and it gives you 70, 70 is what the big part is and 70+20 is 90, which is the size of a 90degree side.

4(20)-10

80-10

70

70+20=90

3 0
3 years ago
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At a certain school,5/9 of the students are boys. Today, 2/3 of the boys brought their lunch. What fraction of the students are
Elodia [21]
5/6 of the boys brought their lunch.



In order to determine the answer you need to divide 5/9 by 2/3 in which will give you 5/6.
4 0
3 years ago
3<br> Solve for x. Round answer to the nearest tenth.<br> X<br> 759<br> 17
QveST [7]

Answer:

x ≈ 4.6

Step-by-step explanation:

Reference angle = 75°

Opposite side to reference angle = 17

Adjacent side = x

Applying TOA:

Tan 75 = opp/adj

Tan 75 = 17/x

Multiply both sides by x

x*Tan 75 = 17

Divide both sides by Tan 75

x = 17/Tan 75

x ≈ 4.6

7 0
3 years ago
A dairy farm uses the somatic cell count (SCC) report on the milk it provides to a processor as one way to monitor the health of
Eva8 [605]

Answer:

t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37  

p_v =P(t_{4}>2.37)=0.038  

If we compare the p value and a significance level assumed \alpha=0.1 we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.  

Step-by-step explanation:

Data given and notation

\bar X=250000 represent the sample mean  

s=37500 represent the standard deviation for the sample

n=5 sample size  

\mu_o =210250 represent the value that we want to test  

\alpha=0.1 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the mean is higher than 210250, the system of hypothesis would be:  

Null hypothesis:\mu \leq 210250  

Alternative hypothesis:\mu > 210250  

Compute the test statistic  

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

We can replace in formula (1) the info given like this:  

t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37  

Now we need to find the degrees of freedom for the t distirbution given by:

df=n-1=5-1=4

Conclusion

Since is a one right tailed test the p value would be:  

p_v =P(t_{4}>2.37)=0.038  

If we compare the p value and a significance level assumed \alpha=0.1 we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.  

6 0
3 years ago
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kkurt [141]

Answer:

34

Step-by-step explanation:

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3 years ago
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