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dem82 [27]
3 years ago
11

5,ooo invested 5 years 3% interest compounded continuously

Mathematics
1 answer:
galben [10]3 years ago
6 0
A=pe^rt

A future value
P principle 5000
R interest rate 3/100 =0.03
T time 5 years

A=5,000×e^(0.03×5)
A=5,809.17
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If 5^2x=26, what is the value of x?<br><br> x≈4.05<br> x≈1.41<br> x≈1.01<br> x≈0.99
babunello [35]
The correct answer is x≈1.01
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7 0
3 years ago
11. Joe bought 3 pairs of jeans for $71.40. a. How much did he pay per pair? b. How much would be need to pay for 8 pairs of jea
Vikki [24]

Answer:

$23.80 per pair, $190.4 for 8 pairs

Step-by-step explanation:

a. assuming each pair of jeans is the same price, just divide the total by 3. the price per pair is 71.40/3, which is $23.80 per pair

b. simply multiply the price of one pair of jeans, as found above, by 8. 23.80*8 is $190.4 for 8 jeans

hope this helped!

4 0
2 years ago
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Scorpion4ik [409]

Answer:

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7 0
3 years ago
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A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. ​a) E
polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) \alpha =1-0.98=0.02

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq  p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and z_{\alpha/2} is the z-value that let a probability of \alpha/2 on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, \alpha by 0.02 and z_{\alpha/2} by 2.33

Where p' and \alpha are calculated as:

p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02

So, replacing the values we get:

0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq  p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence 1-\alpha is equal to 98%

It means the the level of significance \alpha is:

\alpha =1-0.98=0.02

4 0
3 years ago
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