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3 years ago

I just need help with 15

Mathematics

1 answer:

Iteru [2.4K]3 years ago

4 0

Answer:

The answer is survey five randomly selected students from every homeroom

Step-by-step explanation:

hope it helps

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Use the explicit formula an = a1 + (n - 1) • d to find the 350th term of the sequence below. 57, 66, 75, 84, 93, ... A. 3234 B.

mylen [45]

To use the equation we need a1 and d(the common difference).
We have a1 = 57, we need to find d.
93-84 = 9; 84-75=9; 75-66=9; 66-57=9
d = 9
a350 = 57 + (350-1)(9)
a350 = 57 + (349)(9)
a350 = 57 + 3141
a350 = 3198

4 0

3 years ago

Read 2 more answers

You decide to put $5000 in a savings account to save $6000 down payment on a new car. If the account has an interest rate of 7%

bezimeni [28]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill&\$6000\\ P=\textit{original amount deposited}\dotfill &\$5000\\ r=rate\to 7\%\to \frac{7}{100}\dotfill &0.07\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years \end{cases}$

$\bf 6000=5000\left(1+\frac{0.07}{12}\right)^{12\cdot t}\implies \cfrac{6000}{5000}\approx (1.0058)^{12t}\implies \cfrac{6}{5}\approx(1.0058)^{12t} \\\\\\ \log\left( \cfrac{6}{5} \right)\approx \log[(1.0058)^{12t}]\implies \log\left( \cfrac{6}{5} \right)\approx 12t\log(1.0058) \\\\\\ \cfrac{\log\left( \frac{6}{5} \right)}{12\log(1.0058)}\approx t\implies 2.63\approx t\impliedby \textit{about 2 years, 7 months and 16 days}$

6 0

3 years ago

Landen has his daily video gaming time each day for 7 days. But one day he plays and extra 50 minutes. His total video playing t

mojhsa [17]

Answer:

im thinking 25 or 24

Step-by-step explanation:

sorry if i made you get it wrong

8 0

3 years ago

A sailor is 30m above the water in the crow's nest on a sailboat. The sailor encounters an orca surface at an angle of depressio

Natali [406]

Given :

A sailor is 30 m above the water in the crow's nest on a sailboat.

The sailor encounters an orca surface at an angle of depression of 15 degrees.

The crows nest is 20 m horizontally from the bow (front) of the boat.

To Find :

How far in front of the boat is the orca.

Solution :

Let, distance of boat front from the crow's nest is x.

So,

$\dfrac{30}{20+x}=tan \ 15^{\circ}\\\\x=\dfrac{30}{tan \ 15^{\circ}}-20\\\\x=111.94-20\\\\x=91.94\ m$

Hence, this is the required solution.

4 0

3 years ago

Joey went to a toy show at the local community center. He bought action figures and trading cards. He spent a total amount of $3

GuDViN [60]

$34.04 - $4.22 = $29.82
$29.82 ÷ 6 = A

3 0

3 years ago