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agasfer [191]
2 years ago
9

PLease help me this is due at midnight!

Mathematics
2 answers:
wel2 years ago
6 0
What do you need help with? Please give me a photo or something so that I can help you with it. All I see is “Please help me, this is due at midnight!!” Next time please give US a picture or something so I can be able to help you, Thank you, Sincerely Unknown.
rusak2 [61]2 years ago
3 0
I have no idea how to do this question or anything I just want to answer for the points so my bad
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2. Harry just deposited $1500 into a savings account giving 6% interest compounded quarterly.a) How much will be in the account
Studentka2010 [4]

The formula for compound interest is:

\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ \text{where,} \\ A=\text{ Final amount} \\ r=\text{ Interest rate} \\ n=\text{ Number of times interest applied per period} \\ t=\text{ Number of time period elapsed} \\ P=\text{ Intial principal balance} \end{gathered}

Given data:

\begin{gathered} P=\text{ \$1500} \\ r=6\text{ \%}=0.06 \\ n=4\text{ times (compounded quarterly)} \end{gathered}

a. After ten years, that is t = 10 years, the amount in the account will be

\begin{gathered} A=1500(1+\frac{0.06}{4})^{4\times10} \\ A=\text{ }1500(1+0.015)^{40} \\ A=\text{ }1500(1.015)^{40} \\ A=\text{ \$2721.03} \end{gathered}

b. After twenty years, that is t = 20 years, the amount in the account will be:

\begin{gathered} A=1500(1+\frac{0.06}{4})^{4\times20} \\ A=1500(1.015)^{4\times20} \\ A=1500(1.015)^{80} \\ A=\text{ \$}4935.99 \end{gathered}

c. The time it takes for Harry's initial account value to double will be:

\begin{gathered} A=2\text{ x initial value = 2 }\times\text{ \$1500 = \$3000} \\ 3000=1500(1.015)^{4t} \\ (1.015)^{4t}=\frac{3000}{1500} \\ (1.015)^{4t}=2 \\ \text{ Find the logarithm of both sides} \\ \ln (1.015)^{4t}=\ln 2 \\ 4t=\frac{\ln 2}{\ln 1.015} \\ 4t=46.56 \\ t=\frac{46.56}{4}=11.64 \end{gathered}

Therefore, the time it takes Harry's initial account to double is approximately 11 years

8 0
1 year ago
a number, half of that number, and one-third of that number are added. The result is 22. What is the original number
aliina [53]
<span>A number, half of that number, and one-third of that number are added.
This statement can be expressed as:
x + (1/2)x + (1/3)x

The total is 22
</span>x + (1/2)x + (1/3)x = 22

So let's look for x
x + (1/2)x + <span>(1/3)x = 22
</span>(6x + 3x + 2x) / 6 = 22
11x = 132
x = 12

So the original number is 12.
7 0
3 years ago
If GHJ is equal to LMK, with a scale factor of 5:6, find the perimeter of GHJ.
xxTIMURxx [149]

Answer:  Weeeell,  35

Step-by-step explanation:

1. By definition, the perimeter of a triangle is the sum of the lenght of each side.

2. Then, the perimeter of the triangle LMK is:

P_{LMK}=14+17+11\\P_{LMK}=42

3. If both triangles are congruent and the scale factor is 5:6 (Which you can rewrite as a fraction: 5/6), you must multiply the perimeter of the triangle   LMK by this scale factor.

4. Then, you have that the perimeter of the triangle GHJ is:

P_{GHJ}=42*\frac{5}{6}=35

<em><u>What I have to say:</u></em>

<u><em>I hope I helped you, and pls mark brainliest. Have a great day BYYYE!!!!!</em></u>

4 0
3 years ago
.In the Star Wars franchise, Yoda stands at only 66 centimeters tall. Suppose you want to see whether or not hobbits from the Lo
r-ruslan [8.4K]

Answer:

We conclude that the average height of hobbit is taller than Yoda.

Step-by-step explanation:

We are given that in the Star Wars franchise, Yoda stands at only 66 centimetres tall.

From a sample of 7 hobbits, you find their mean height \bar X = 80 cm with standard deviation s = 10.8 cm.

<u><em /></u>

<u><em>Let </em></u>\mu<u><em> = average height of hobbit.</em></u>

So, Null Hypothesis, H_0 : \mu \leq 66 cm      {means that the average height of hobbit is shorter than or equal to Yoda}

Alternate Hypothesis, H_A : \mu > 66 cm      {means that the average height of hobbit is taller than Yoda}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about population standard deviation;

                     T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}  ~ t_n_-_1

where, \bar X = sample mean height = 80 cm

             s = sample standard deviation = 10.8 cm

             n = sample of hobbits = 7

So, <u><em>test statistics</em></u>  =  \frac{80-66}{\frac{10.8}{\sqrt{7}}}  ~ t_6

                              =  3.429

The value of t test statistics is 3.429.

<em>Now, at 0.01 significance level the t table gives critical value of 3.143 for right-tailed test. Since our test statistics is more than the critical value of t as 3.429 > 3.143, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>

<em />

Therefore, we conclude that the average height of hobbit is taller than Yoda.

3 0
3 years ago
-4.7 divided by 1\4 ????????????????
Nady [450]

Answer:

-18.8

Step-by-step explanation:

1/4 as a decimal is 0.25 so then if you divide -4.7 by 0.25 the answer is -18.8 :)

I hope this helped :)

3 0
3 years ago
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