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oksian1 [2.3K]
2 years ago
14

Jazelle Momba wants to visit her family in Zimbabwe in 2025, which is 6 years from now. She knows that it will cost approximatel

y $8,000 including flight costs, on-the-ground costs, and extra spending money to stay for 4 months. If she opens an account that compounds interest at 4% semiannually, how much does she need to deposit today to cover the total cost of her visit?
Mathematics
1 answer:
Ludmilka [50]2 years ago
8 0

9514 1404 393

Answer:

  $6307.95

Step-by-step explanation:

The compound interest formula can help with that.

  A = P(1 +r/n)^(nt) . . . . value of principal P at rate r for t years, compounded n times per year.

  P = A(1 +r/n)^(-nt) = $8000(1 +0.04/2)^(-2·6) = $8000(1.02^-12) = $6307.95

Momba needs to deposit $6307.95 today to have $8000 in 6 years.

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A toy American Eskimo dog has a mean weight of 8 pounds with a standard deviation of 1 pound. Assuming the weights of toy Eskimo
Bumek [7]

Answer:

Option b) 6-10 lbs

Step-by-step explanation:

Given that X, the weight of  toy American Eskimo dog has mean of 8 pounds with a standard deviation of 1 pound

For 95% of the dogs according to normal distribution would lie between 2 std deviations on either side of the mean.

i.e. lower bound= Mean- 2 std deviation = 8-2 =6

Upper bound = Mean +2 std dev = 8+2 =10

Hence range of weights would be

6-10 lbs

Option b is the answer

6 0
3 years ago
What is the value of 25 – 5 × 8 – 4?
Galina-37 [17]

Answer:

25- 40 - 4 = -19

8 0
3 years ago
Read 2 more answers
How do I solve this?
krok68 [10]
3 is 95 and 4 is 85. I believe
4 0
3 years ago
Help me please on these questions
Fittoniya [83]
The answer to number 23 is D, 49.
8 0
3 years ago
The calibration of a scale is to be checked by weighing a 13 kg test specimen 25 times. Suppose that the results of different we
Natali5045456 [20]

Solution :

a).

Given : Number of times, n = 25

            Sigma, σ = 0.200 kg

            Weight, μ = 13 kg

Therefore the hypothesis should be tested are :

$H_0 : \mu = 13 $

$H_a : \mu \neq 13$

b). When the value of $\overline x = 12.84$

 Test statics :

   $Z=\frac{(\overline x - \mu)}{\frac{\sigma}{\sqrt n}} $

  $Z=\frac{(14.82-13)}{\frac{0.2}{\sqrt {25}}} $

          $=\frac{1.82}{0.04}$

          = 45.5  

P-value = 2 x P(Z > 45.5)

            = 2 x 1 -P (Z < 45.5) = 0

Reject the null hypothesis if P value  < α = 0.01 level of significance.

So reject the null hypothesis.

Therefore, we conclude that the true mean measured weight differs from 13 kg.

3 0
3 years ago
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