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Alenkinab [10]
3 years ago
14

A young college student puts $2,500 in savings into a special bank account. The account earns interest at a rate of 3%, compound

ed 6 times per year. The bank account matures after 2 years, when the college student may withdraw the funds. How much money will the account have at maturation?
Mathematics
1 answer:
Serggg [28]3 years ago
3 0
<span>2500(1 + .03/6)^12 = 2654.19
Your answer is $2654.2
Hope this helps!</span>
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Of 30 students, 1/3 play sports. Of those who play sports, 2/5 play soccer.
lana [24]

Answer: 10 or 12

Step-by-step explanation:

6 0
3 years ago
Help me with the first step
grandymaker [24]
First find the area of the triangle: (4 x 6) ÷ 2 = 12
Then the area of the rectangle: 6 x 8 = 48
Last combine: 48 + 12 = 60

Lol, sorry~

Hope this helped tho
4 0
3 years ago
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
2 years ago
Braden has $75 to spend for t shirts online. If shipping is $3 , how many " t" t shirts can he buy if theshirts cost $ 8?
inysia [295]

Answer:

9 tshirts

Step-by-step explanation:

Money he had for t shirts= $75

Cost of shipping= $3

Money left for t shirts= $75-$3=$72

Cost of 1 t shirt= $8

We can find the answer with two ways:

1st way: tshirts of $8=1

tshirts of $72=72/8

=$9

2nd way: Cost ($) 8 72

T shirts 1 x

8/1=72/x

by cross multiply

x=72/8

x=9

7 0
2 years ago
Type SSS, SAS, ASA, SAA, or HL to<br> describe these triangles.
Viktor [21]

Answer:

ASA

Step-by-step explanation:

none needed [=

3 0
2 years ago
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