Please really need help on this please

How many feet is 20% of 4 miles

Nataliya [291]

1 mile=5280 feet
4 miles=21120 feet

21120×20÷100=4224. As a result, 20% of 4 miles is 4224 feet. Hope it help!

3 0

3 years ago

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Using the Law of Cosines, in triangle DEF, if e=18yd, d=10yd, f=22yd, find measurement of angle D

Ivahew [28]

Check the picture below.

$\bf \textit{Law of Cosines}\\ \quad \\
c^2 = {{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)\implies 
c = \sqrt{{{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)}
\\\\\\
\cfrac{{{ a}}^2+{{ b}}^2-c^2}{2{{ a}}{{ b}}}=cos(C)\implies cos^{-1}\left(\cfrac{{{ a}}^2+{{ b}}^2-c^2}{2{{ a}}{{ b}}}\right)=\measuredangle C\\\\
-------------------------------\\\\
\begin{cases}
d=10\\
e=18\\
f=22
\end{cases}\implies cos^{-1}\left(\cfrac{{{ 18}}^2+{{ 22}}^2-10^2}{2(18)(22)}\right)=\measuredangle D$

$\bf \implies cos^{-1}\left(0.893\overline{93}\right)=\measuredangle D\implies 26.63^o\approx \measuredangle D$

7 0

3 years ago

The amount of time that people spend at Grover Hot Springs is normally distributed with a mean of 73 minutes and a standard devi

Vesnalui [34]

Answer:

(a) $X\sim N(\mu = 73, \sigma = 16)$

(b) 0.7910

(c) 0.0401

(d) 0.6464

Step-by-step explanation:

Let X = amount of time that people spend at Grover Hot Springs.

The random variable X is normally distributed with a mean of 73 minutes and a standard deviation of 16 minutes.

(a)

The distribution of the random variable X is:

$X\sim N(\mu = 73, \sigma = 16)$

(b)

Compute the probability that a randomly selected person at the hot springs stays longer than 60 minutes as follows:

$P(X>60)=P(\frac{X-\mu}{\sigma}>\frac{60-73}{16})\\=P(Z>-0.8125)\\=P(Z$

*Use a z-table for the probability.

Thus, the probability that a randomly selected person at the hot springs stays longer than an hour is 0.7910.

(c)

Compute the probability that a randomly selected person at the hot springs stays less than 45 minutes as follows:

$P(X$

*Use a z-table for the probability.

Thus, the probability that a randomly selected person at the hot springs stays less than 45 minutes is 0.0401.

(d)

Compute the probability that a randomly person spends between 60 and 90 minutes at the hot springs as follows:

$P(60$

*Use a z-table for the probability.

Thus, the probability that a randomly person spends between 60 and 90 minutes at the hot springs is 0.6464

6 0

3 years ago

-(-6x - 5) = 4(5x + 3)

Lina20 [59]

Answer:

$\large\boxed{B)\ -\dfrac{1}{2}}$

Step-by-step explanation:

$-(-6x-5)=4(5x+3)\qquad\text{use the distributive property}\\\\-(-6x)-(-5)=(4)(5x)+(4)(3)\\\\6x+5=20x+12\qquad\text{subtrac 5 from both sides}\\\\6x=20x+7\qquad\text{subtract 20x from both sides}\\\\-14x=7\qquad\text{divide both sides by (-14)}\\\\x=-\dfrac{7}{14}\\\\x=-\dfrac{1}{2}$

8 0

3 years ago

Anise wants to have longer hair than her sister. Her sister's hair is 2.25 feet long. How long does Anise's hair have to

rodikova [14]

The answer has to 2.26 because 2.26 is bigger than 2.25

3 0

2 years ago

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