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2 years ago

Please help! I am having trouble with this lesson.

Mathematics

2 answers:

jeka57 [31]2 years ago

8 0

Answer:

Step-by-step explanation:

1 3/4 ÷1/6= 21/2 or 10 1/2

2/5 ÷ 1/10= 4

1 3/4 ÷1/10= 35/2 or 17 1/2 <---- greatest value

2/5 ÷1/6=12/5 or 2 2/5

tia_tia [17]2 years ago

7 0

here i solved it on paper

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How many terms will the following expression have once it is simplified?

MA_775_DIABLO [31]

Answer:

Step-by-step explanation:

Let's multiply this out and combine like terms:

$3x(x+1)+x\\=3x^2+3x+x\\=3x^2+4x$

So there are 2 terms.

Answer choice B is right.

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3 years ago

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What is 5-2+5-54-3+405+12-9

Ad libitum [116K]

Answer:

359

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5-2+5-54-3+405+12-9=359

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2 years ago

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System of equation x 3y=9 y=x-1

olga2289 [7]

x + 3y = 9
-x + y = -1
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3 years ago

You need a mean of at least 90 points to advance to the next round of the touch-screen trivia. What score in the fifth game will

lianna [129]

Answer:

At least 98 is needed in the 5th game

Step-by-step explanation:

The missing parameters are:

$Game\ 1 = 95$

$Game\ 2 = 91$

$Game\ 3 = 77$

$Game\ 4 =89$

$Mean = 90$ at least

Required

The score in game 5 to make you advance

Mean is calculated as:

$Mean = \frac{\sum x}{n}$

So, we have:

$Mean = \frac{95 + 91 + 77 + 89 + Game\ 5}{5}$

$Mean = \frac{352 + Game\ 5}{5}$

The mean must be at least 90.

So, we have:

$\frac{352 + Game\ 5}{5} \ge 90$

Multiply both sides by 5

$5 * \frac{352 + Game\ 5}{5} \ge 90 * 5$

$352 + Game\ 5 \ge 450$

Make Game 5 the subject

$Game\ 5 \ge450 - 352$

$Game\ 5 \ge 98$

At least 98 is needed in the 5th game

4 0

3 years ago

The level of nitrogen oxides (nox) in the exhaust after 50,000 miles or fewer of driving of cars of a particular model varies no

mina [271]

Answer: 0.533

Explanation: Note that

$x \ \textgreater \ I \\ \Leftrightarrow x - 0.3 \ \textgreater \ I - 0.3 \\ \\ \Leftrightarrow \boxed{\frac{x - 0.3}{0.1} \ \textgreater \ \frac{I - 0.3}{0.1}} (1)$

Let

$z_1$ = z-score for I.
$z_2$ = z-score for x.

The formula for z-score is given by

$\text{z-score} = \frac{(\text{data point}) - (\text{mean})}{(\text{standard deviation})}$

So,

$z_2 = \frac{x - 0.3}{0.1}$

$z_1 = \frac{I - 0.3}{0.1}$

Based on inequality in (1),

$z_2 > \ z_1$

So,

$P(x \ \textgreater \ I) = 0.01
\\
\\ \Leftrightarrow P \left (\frac{x - 0.3}{0.1} \ \textgreater \ \frac{I - 0.3}{0.1} \right ) = 0.01 
\\
\\ \Leftrightarrow P(z_2 \ \textgreater \ z_1) = 0.01
\\
\\ \Leftrightarrow 1 - P(z_2 \leq z_1) = 0.01
\\
\\ \Leftrightarrow \boxed{P(z_2 \leq z_1) = 0.99}$

Since $z_1$ and $z_2$ are z-scores and the level of nitrogen oxides are normally distributed, using normal distribution calculator (or table),

$z_1 = 2.326
\\
\\ \Leftrightarrow \frac{I - 0.3}{0.1} = 2.326
\\
\\ \Leftrightarrow I - 0.3 = 0.1(2.326)
\\
\\ \Leftrightarrow I - 0.03 = 0.2326
\\
\\ \Leftrightarrow \boxed{I \approx 0.533}$

4 0

3 years ago