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3 years ago

3.

1 answer:

5 0

Correct Answer: Option A

Principal Amount = P = $23,400
Interest rate = r = 3% = 0.03
Time = t = 10
Number of compounding periods in a year = n = 2
Compounded amount = A = ?

Formula for compound interest is:

$A = P(1+ \frac{r}{n} )^{t*n}$

Using the values, we get:

$A=23400(1+ \frac{0.03}{2})^{2*10}=31516.41$

Option A is the closest one, so its the correct answer

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25% of all who enters a race do not complete. 30 haveentered.

nikklg [1K]

Answer:

The probability that exactly 5 are unable to complete the race is 0.1047

Step-by-step explanation:

We are given that 25% of all who enters a race do not complete.

30 have entered.

what is the probability that exactly 5 are unable to complete the race?

So, We will use binomial

Formula : $P(X=r) =^nC_r p^r q^{n-r}$

p is the probability of success i.e. 25% = 0.25

q is the probability of failure = 1- p = 1-0.25 = 0.75

We are supposed to find the probability that exactly 5 are unable to complete the race

n = 30

r = 5

$P(X=5) =^{30}C_5 (0.25)^5 (0.75)^{30-5}$

$P(X=5) =\frac{30!}{5!(30-5)!} \times(0.25)^5 (0.75)^{30-5}$

$P(X=5) =0.1047$

Hence the probability that exactly 5 are unable to complete the race is 0.1047

6 0

4 years ago

A certain brand of coffee comes in two sizes. An 11.3-ounce package costs $2.54. A 27.8 ounce costs $6.78. Find the unit price f

Sauron [17]

Step-by-step explanation:

for the 11.3 ounce one:

1 ounce is $0.2247787611=$0.22

for the 27.8 ounce one:

1 ounce is $0.2438848921=$0.24

the 11.3 ounce one is a better buy as 1 ounce of it cost lesser than the 1 ounce of 27.8 ounce package.

3 0

4 years ago

Land's Bend sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the intern

Naya [18.7K]

Answer:

80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases ( $\mu_1-\mu_2$ ) is [-9.132 , 23.332].

Step-by-step explanation:

We are given that a random sample of 7 sales receipts for mail-order sales results in a mean sale amount of $81.70 with a standard deviation of $18.75.

A random sample of 11 sales receipts for internet sales results in a mean sale amount of $74.60 with a standard deviation of $28.25.

Firstly, the Pivotal quantity for 80% confidence interval for the difference between population means is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean sales receipts for mail-order sales = $81.70

$\bar X_2$ = sample mean sales receipts for internet sales = $74.60

$s_1$ = sample standard deviation for mail-order sales = $18.75

$s_2$ = sample standard deviation for internet sales = $28.25

$n_1$ = size of sales receipts for mail-order sales = 7

$n_2$ = size of sales receipts for internet sales = 11

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(7-1)\times 18.75^{2} +(11-1)\times 28.25^{2} }{7+11-2} }$ = 25.11

Here for constructing 80% confidence interval we have used Two-sample t test statistics as we don't know about population standard deviations.

So, 80% confidence interval for the difference between population means, ( $\mu_1-\mu_2$ ) is ;

P(-1.337 < $t_1_6$ < 1.337) = 0.80 {As the critical value of t at 16 degree

of freedom are -1.337 & 1.337 with P = 10%}

P(-1.337 < $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 1.337) = 0.80

P( $-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ${(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}$ < $1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.80

P( $(\bar X_1-\bar X_2)-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.80

80% confidence interval for ( $\mu_1-\mu_2$ ) =

[ $(\bar X_1-\bar X_2)-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ]

= [ $(81.70-74.60)-1.337 \times {25.11 \times \sqrt{\frac{1}{7} +\frac{1}{11} } }$ , $(81.70-74.60)+1.337 \times {25.11 \times \sqrt{\frac{1}{7} +\frac{1}{11} } }$ ]

= [-9.132 , 23.332]

Therefore, 80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases ( $\mu_1-\mu_2$ ) is [-9.132 , 23.332].

4 0

3 years ago

PLEASE SOMEONE HELP ME I WILL GIVE BRAINLIEST

Sidana [21]

20 a: x = 5

20 b: x = 12

17 c: 30

8 0

3 years ago

WHich system of inequalities is shown?

Tom [10]

Answer:

its one of the 2 thats in the middle

Step-by-step explanation:

6 0

3 years ago