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3 years ago

0.139 times what number could get me to 100? PLEASE ANSWER, FAST

Mathematics

1 answer:

Triss [41]3 years ago

7 0

Answer:

719.4244604

Step-by-step explanation:

100/0.139=719.4244604

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Ryan had a balance of −$767.25 in his bank account, and Tim had a balance of −$556.75 in his bank account. Which statement expla

kipiarov [429]

Answer: -767.25<-556.75

Explanation: Ryan has less money than tim therefore making him owe the bank more

8 0

3 years ago

Read 2 more answers

I really don’t get this at all. kinda need help

rewona [7]

Same . It’s not giving me a lot of information!!

5 0

3 years ago

Q4.

goblinko [34]

The coding of the statistic is used to make it easier to work with the large sunshine data set

The mean of the sunshine is 3.05 $\overline 6$

The standard deviation is approximately 18.184

Reason:

The given parameters are;

The sample size, n = 3.

∑x = 947

Sample corrected sum of squares, Sₓₓ = 33,065.37

The mean and standard deviation = Required

Solution:

$Mean, \ \overline x = \dfrac{\sum x_i}{n}$

The mean of the daily total sunshine is therefore;

$Mean, \ \overline x = \dfrac{947}{30} \approx 31.5 \overline 6$

$s = \dfrac{x}{10 } - \dfrac{1}{10}$

$E(s) = \dfrac{Ex}{10 } - \dfrac{1}{10}$

$E(s) = \dfrac{31.5 \overline 6}{10 } - \dfrac{1}{10} = 3.05 \overline 6$

The mean ≈ 3.05 $\overline 6$

Alternatively

, $The \ mean \ of \ the \ daily \ total \ sunshine, \, s = \dfrac{31.5 \overline 6 - 1}{10 } = 3.05\overline 6$

The mean of the daily total sunshine, $\overline s$ ≈3.05 $\overline 6$

$Var(s) = Var \left(\dfrac{x}{10 } - \dfrac{1}{10} \right)$

$Var(s) = \left(\dfrac{1}{10}\right)^2 \times Var \left(x \right)$

Therefore;

$Var(s) = \left(\dfrac{1}{10}\right)^2 \times 33,065.37 = 330,6537$

Therefore;

$s = \sqrt{330.6537} \approx 18.184$

The standard deviation, $s_s$ ≈ 18.184

Learn more about coding of statistic data here:

brainly.com/question/14837870

3 0

2 years ago

Will give brainliest!

Gala2k [10]

Check the picture below, it hits the ground when y = 0.

$\bf ~~~~~~\textit{initial velocity}\\\\
\begin{array}{llll}
~~~~~~\textit{in feet}\\\\
h(t) = -16t^2+v_ot+h_o \\\\
\end{array} 
\quad 
\begin{cases}
v_o=\stackrel{0}{\textit{initial velocity of the object}}\\\\
h_o=\stackrel{162}{\textit{initial height of the object}}\\\\
h=\stackrel{}{\textit{height of the object at "t" seconds}}
\end{cases}
\\\\\\
h(t)=-16t^2+0t+162\implies 0=-16t^2+162\implies 16t^2=162
\\\\\\
t=\cfrac{162}{16}\implies t=\cfrac{81}{8}\implies t=10\frac{1}{8}$

6 0

3 years ago

Write an equation of the line that passes through (-2,3) and is parallel to the line y = 2x – 4.

allochka39001 [22]

Answer:

The parallel line should be y = 2x + 7

3 0

3 years ago