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Doss [256]
3 years ago
15

HELP PLS WILL GIVE BRAINLISTE

Mathematics
2 answers:
kifflom [539]3 years ago
6 0

Answer: C and H

Step-by-step explanation:

plugging in for each equation for the first one C is the only one that comes up with a whole number and with H you divide and then add to get the answer :)

valentina_108 [34]3 years ago
5 0

Answer:

C.

Step-by-step explanation:

all you have to do is replace the x with 14 until the equation is true.

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PLEASEEEEE HELP ASAP!!!
Ne4ueva [31]

Answer:

Step-by-step explanation:

Triangle ABC is a right angle triangle.

From the given right angle triangle

AB represents the hypotenuse of the right angle triangle.

With m∠A as the reference angle,

AC represents the adjacent side of the right angle triangle.

BC represents the opposite side of the right angle triangle.

To determine AB, we would apply the Cosine trigonometric ratio

Cos θ = opposite side/hypotenuse. Therefore,

Cos 52 = 10/AB

0.616 = 10/AB

AB = 10/0.616

AB = 16.23

6 0
3 years ago
Given: PSTK is a trapezoid, m∠P=90° SK =13, PK = 12, ST=8 Find: The area of PSTK.
guajiro [1.7K]
<h3>Given</h3>

trapezoid PSTK with ∠P=90°, KS = 13, KP = 12, ST = 8

<h3>Find</h3>

the area of PSTK

<h3>Solution</h3>

It helps to draw a diagram.

∆ KPS is a right triangle with hypotenuse 13 and leg 12. Then the other leg (PS) is given by the Pythagorean theorem as

... KS² = PS² + KP²

... 13² = PS² + 12²

... PS = √(169 -144) = 5

This is the height of the trapezoid, which has bases 12 and 8. Then the area of the trapezoid is

... A = (1/2)(b1 +b2)h

... A = (1/2)(12 +8)·5

... A = 50

The area of trapezoid PSTK is 50 square units.

5 0
3 years ago
Businesses deposit large sums of money into bank accountsImagine an account with $10 million dollars in it.
adoni [48]

again, let's assume daily compounding means 365 days per year.

~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$10000000\\ r=rate\to 2.12\%\to \frac{2.12}{100}\dotfill &0.0212\\ t=years\dotfill &1 \end{cases} \\\\\\ I = (10000000)(0.0212)(1)\implies \boxed{I=212000}

~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$10000000\\ r=rate\to 2.12\%\to \frac{2.12}{100}\dotfill &0.0212\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{daily, thus 365} \end{array}\dotfill &365\\ t=years\dotfill &1 \end{cases}

A=10000000\left(1+\frac{0.0212}{365}\right)^{365\cdot 1}\implies A\approx 10214256.88 \\\\\\ \underset{\textit{earned interest amount}}{10214256.88~~ - ~~10000000 ~~ \approx ~~ \boxed{214256.88}}

what's their difference?  well

\stackrel{\textit{compounded daily}}{\approx 214256.88}~~ - ~~\stackrel{\textit{simple interest}}{212000}\implies \boxed{2256.88}

7 0
2 years ago
Which expression is equivalent to x y Superscript two-ninths?
crimeas [40]

Option D: x\sqrt[9]{y^2} is the expression equivalent to xy^{\frac{2}{9}}

Explanation:

Option A: \sqrt{xy^9}

The expression can be written as ({xy^9})^{\frac{1}{2}

Applying exponent rule, we get,

x^{\frac{1}{2}} y^{\frac{9}{2}}

Thus, the expression \sqrt{xy^9} is not equivalent to the expression xy^{\frac{2}{9}}

Hence, Option A is not the correct answer.

Option B: \sqrt[9]{xy^2}

The expression can be written as ({xy^2})^{\frac{1}{9}

Applying exponent rule, we get,

x^{\frac{1}{9}} y^{\frac{2}{9}}

Thus, the expression \sqrt[9]{xy^2} is not equivalent to the expression xy^{\frac{2}{9}}

Hence, Option B is not the correct answer.

Option C: x\sqrt{y^9}

The expression can be written as x(y^9)^{\frac{1}{2} }

Applying exponent rule, we get,

x y^{\frac{9}{2}}

Thus, the expression x\sqrt{y^9} is not equivalent to the expression xy^{\frac{2}{9}}

Hence, Option C is not the correct answer.

Option D: x\sqrt[9]{y^{2} }

The expression can be written as x(y^2)^{\frac{1}{9} }

Applying exponent rule, we get,

xy^{\frac{2}{9}}

Thus, the expression xy^{\frac{2}{9}} is equivalent to the expression xy^{\frac{2}{9}}

Hence, Option D is the correct answer.

4 0
3 years ago
Read 2 more answers
The number of major earthquakes in a year is approximately normally distributed with a mean of 20.8 and a standard deviation of
SCORPION-xisa [38]

Answer:

a) 51.60% probability that in a given year there will be less than 21 earthquakes.

b) 49.35% probability that in a given year there will be between 18 and 23 earthquakes.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 20.8, \sigma = 4.5

a) Find the probability that in a given year there will be less than 21 earthquakes.

This is the pvalue of Z when X = 21. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{21 - 20.8}{4.5}

Z = 0.04

Z = 0.04 has a pvalue of 0.5160.

So there is a 51.60% probability that in a given year there will be less than 21 earthquakes.

b) Find the probability that in a given year there will be between 18 and 23 earthquakes.

This is the pvalue of Z when X = 23 subtracted by the pvalue of Z when X = 18. So:

X = 23

Z = \frac{X - \mu}{\sigma}

Z = \frac{23 - 20.8}{4.5}

Z = 0.71

Z = 0.71 has a pvalue of 0.7611

X = 18

Z = \frac{X - \mu}{\sigma}

Z = \frac{18 - 20.8}{4.5}

Z = -0.62

Z = -0.62 has a pvalue of 0.2676

So there is a 0.7611 - 0.2676 = 0.4935 = 49.35% probability that in a given year there will be between 18 and 23 earthquakes.

5 0
3 years ago
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