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lozanna [386]
3 years ago
15

Can someone tell me what 2/3 divided by 5 is it’s for a quiz.

Mathematics
2 answers:
elena-14-01-66 [18.8K]3 years ago
8 0
It’s simple just do stay switch flip 1. Keep 2/3 2. Make division to multiplication then...3. Turn 5 into 1/5! Then you get 2/15! The question might ask you to simplify the fraction.
docker41 [41]3 years ago
7 0

Answer:2/15 hope this helps

Step-by-step explanation:

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According to a human modeling​ project, the distribution of foot lengths of women is approximately Normal with a mean of 23.4 ce
stiv31 [10]

Answer: 22.0.6%

Step-by-step explanation:

Given : According to a human modeling​ project, the distribution of foot lengths of women is approximately Normal with \mu=23.3\ cm and \sigma=1.3\ cm.

In the United​ States, a​ woman's shoe size of 6 fits feet that are 22.4 centimeters long.

Then, the probability that women in the United States will wear a size 6 or​ smaller :-

P(x\leq22.4)=P(z\leq\dfrac{22.4-23.4}{1.3})\ \ [\because z=\dfrac{x-\mu}{\sigma}]\\\\\approx P(z\leq-0.77)\\\\=1-P(z\leq0.77)\\\\=1-0.77935=0.2206499\approx0.2206=22.06\%

Hence,  the required answer = 22.0.6%

4 0
3 years ago
18. The sum of 'l6, /3, and 'Is is
lys-0071 [83]

Answer:

wn work

Step-by-step explanation:

3 0
3 years ago
The midpoint of A (-4, 2) and B(8, 5) is
Norma-Jean [14]
\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ -4}}\quad ,&{{ 2}})\quad 
%  (c,d)
&({{ 8}}\quad ,&{{ 5}})
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left(\cfrac{{{ 8}} -4}{2}\quad ,\quad \cfrac{{{ 5}} + {{ 2}}}{2} \right)
8 0
3 years ago
Read 2 more answers
The annual precipitation amounts in a certain mountain range are normally distributed with a mean of 91 inches, and a standard d
Vesna [10]
The sample std. dev. will be (14 inches) / sqrt(49), or (14 inches) / 7, or 2 inches.

Find the z score for 93.8 inches:
       93.8 inches - 91.0 inches          2.8 inches
z = ------------------------------------- = ----------------- = 1.4
                    2 inches                         2 inches

Now find the area under the standard normal curve to the left of z = +1.4.

My calculator returns the following:

normalcdf(-100,1.4) = 0.919.  This is the probability that the mean annual precipitation during those 49 years will be less than 93.8 inches.
 
4 0
3 years ago
Please help, basic Algebra question
Tju [1.3M]

Answer:

The given expression  (\frac{5}{6} a^{9}p^{5})  ^{3} = \frac{125}{216}  \times a^{(27) } \times p^{(15)}

Step-by-step explanation:

Here, the given expression is:  (\frac{5}{6} a^{9}p^{5})  ^{3}

Now, starting from the outer most bracket.

As we know :

(abc)^{n}   = (a)^{n} \times (b)^{n}  \times (c)^{n}

and (a^m)^{n} = a^{(m \times n)}

⇒ (\frac{5}{6} a^{9}p^{5})  ^{3} = (\frac{5}{6})^{3} \times (a^{9})^{3}   \times (p^{5}) ^{3}\\

=\frac{125}{216}  \times (a)^{(9\times3) } \times (p)^{(5 \times 3)}\\= \frac{125}{216}  \times a^{(27) } \times p^{(15)}

Hence, the given expression  (\frac{5}{6} a^{9}p^{5})  ^{3} = \frac{125}{216}  \times a^{(27) } \times p^{(15)}

7 0
3 years ago
Read 2 more answers
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