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2 years ago

Mrs. Olga’s $250 investment earns 19% interest compounded annually. What will her balance be after 3 years?

Mathematics

2 answers:

VikaD [51]2 years ago

8 0

Answer:

easports

Step-by-step explanation:

its in the game

Maurinko [17]2 years ago

6 0

Answer:

The answer is A. $421.29

Step-by-step explanation:

Just keep finding 19% of the numbers next, and you add that all up and you get the answer.

250 x 19%=47.5. (250+47.5)=> 297.5x19%= 56.525. (297.5+56.525) =>354.025x19%=67.26475

354.025+67.26475=> 421.28975, which can round up to 421.29.

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Given triangle GHJ, the measure of angle G equals 110°, the measure of angle J equals 40°, and the measure of angle H equals 30°

tester [92]

Answer:

Since angle G is

✔ the largest

angle, the opposite side, JH, is

✔ the longest side

. The order of the side lengths from longest to shortest is

✔ HJ, GH, and GJ

Step-by-step explanation:

5 0

1 year ago

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A rectangle that has an area of 357 square inches is 17 inches wide. Using the formula for area (4072-01-01-03-00_files/i0260000

Masja [62]

A = L * W
A = 357
W = 17

357 = 17L
357 / 17 = L
21 = L <== length is 21 inches

8 0

2 years ago

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Damien , Jake , and Makayla shared a bowl of grapes . Damien ate x grapes . Jake ate 1 more than twice the number of grapes that

vivado [14]

Answer:

There were (4x + 2) grapes in the bowl

Step-by-step explanation:

Here, we are interested in calculating the total number of grapes in the bowl.

We have 3 people sharing the total

Damien are x grapes

Jake ate 1 more than twice what Damien and that is (1 + 2x) grapes

Makayla ate 5 fewer grapes than Damien: So what Makayla ate is (x-5) grapes

And now, we have 6 grapes left.

To find the total number of grapes in the bowl, we need to add up all what they ate plus what is left.

Mathematically, that would be;

x + 1 + 2x + x -5 + 6

= 4x + 2

8 0

2 years ago

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Li wants to add these items to her suitcase: a hairdryer (1.25 lb), a hand-held video game (0.6 lb), an extra video game (0.25 l

lbvjy [14]

Add some of them or all of them to your sum of 47.75, if either or exceeds the limit then that is what left out.

7 0

2 years ago

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A study was performed on green sea turtles inhabiting a certain area. Time-depth recorders were deployed on 6 of the 76 capture

polet [3.4K]

Answer:

One can be 99% confident the true mean shell length lies within the above interval.

The population has a relative frequency distribution that is approximately normal.

Step-by-step explanation:

We are given that Time-depth recorders were deployed on 6 of the 76 captured turtles. These 6 turtles had a mean shell length of 51.3 cm and a standard deviation of 6.6 cm.

The pivotal quantity for a 99% confidence interval for the true mean shell length is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean shell length = 51.3 cm

s = sample standard deviation = 6.6 cm

n = sample of turtles = 6

$\mu$ = true mean shell length

Now, the 99% confidence interval for $\mu$ = $\bar X \pm t_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }$

Here, $\alpha$ = 1% so $(\frac{\alpha}{2})$ = 0.5%. So, the critical value of t at 0.5% significance level and 5 (6-1) degree of freedom is 4.032.

<u>So, 99% confidence interval for</u> $\mu$ = $51.3 \pm 4.032 \times \frac{6.6}{\sqrt{6} }$

= [51.3 - 10.864 , 51.3 + 10.864]

= [40.44 cm, 62.16 cm]

The interpretation of the above result is that we are 99% confident that the true mean shell length lie within the above interval of [40.44 cm, 62.16 cm].

The assumption about the distribution of shell lengths must be true in order for the confidence interval, part a, to be valid is that;

C. The population has a relative frequency distribution that is approximately normal.

This assumption is reasonably satisfied as the data comes from the whole 76 turtles and also we don't know about population standard deviation.

3 0

1 year ago