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2 years ago

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]

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
x + (1/2)x + (1/3)x = 22

So let's look for x
x + (1/2)x + (1/3)x = 22
(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

What I have to say:

I hope I helped you, and pls mark brainliest. Have a great day BYYYE!!!!!

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.

Let $\mu$ = average height of hobbit.

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 One-sample t test statistics 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, test statistics = $\frac{80-66}{\frac{10.8}{\sqrt{7}}}$ ~ $t_6$

= 3.429

The value of t test statistics is 3.429.

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 we reject our null hypothesis.

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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