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7 months ago

This is difficult for me pls help out

Mathematics

1 answer:

Cloud [144]7 months ago

8 0

Answer:

(-0.4)(-2)+(-0.4)(-9)

Step-by-step explanation:

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John found the diagram above in a section about animal tracks in his hiking manual. The pattern of the porcupine tracks shown ca

worty [1.4K]

The answer is C) glide and reflection because if it were graphed, it would be reflected a line equal to x and slid, or glided upward. Hope this helps! :D

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2 years ago

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if you took 7 hours to mow 4 lawns, then at that rate, how many lawns could be lawns could be mowed in 35 hours? at what rate we

liraira [26]

It takes 7 hours to mow 4 lawns
Therefore every hour, (4 ÷ 7) or 4/7 lawns are mowed. This gives the rate.

For 35 hours, 4/7 x 35 lawns are mowed.
=35 ÷ 7 = 5
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3 years ago

What is the median number of runs scored?

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Answer:

i think its 1.5

Step-by-step explanation:

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2 years ago

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Geraldine is graphing the inequality x<6. Which statements below describe steps she may take to graph the inequality correctl

Vlad1618 [11]

Answer:

Draw an open circle at 6

Shade all numbers less than 6

Shade all numbers to the left of 6

Step-by-step explanation:

we have

$x$

The solution of the inequality is the interval (-∞,6)

All real numbers less than 6

In a number line the solution is the shaded area to the left at number 6 (open circle)

see the attached figure to better understand the problem

therefore

Draw an open circle at 6

Shade all numbers less than 6

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3 0

3 years ago

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A bag contains red marbles, white marbles, and blue marbles. Randomly choose two marbles, one at a time, and without replacement

dsp73

Answer:

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

$P(Same) = \frac{67}{210}$

Step-by-step explanation:

Given (Omitted from the question)

$Red = 7$

$White = 9$

$Blue = 5$

Solving (a): $P(First\ White\ and\ Second\ Blue)$

This is calculated using:

$P(First\ White\ and\ Second\ Blue) = P(White) * P(Blue)$

$P(First\ White\ and\ Second\ Blue) = \frac{n(White)}{Total} * \frac{n(Blue)}{Total - 1}$

<em>We used Total - 1 because it is a probability without replacement</em>

So, we have:

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{21 - 1}$

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{20}$

$P(First\ White\ and\ Second\ Blue) = \frac{9*5}{21*20}$

$P(First\ White\ and\ Second\ Blue) = \frac{45}{420}$

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

Solving (b) $P(Same)$

This is calculated as:

$P(Same) = P(First\ Blue\ and Second\ Blue)\$ $or\ P(First\ Red\ and Second\ Red)\ or\ P(First\ White\ and Second\ White)$

$P(Same) = (\frac{n(Blue)}{Total} * \frac{n(Blue)-1}{Total-1})+(\frac{n(Red)}{Total} * \frac{n(Red)-1}{Total-1})+(\frac{n(White)}{Total} * \frac{n(White)-1}{Total-1})$

$P(Same) = (\frac{5}{21} * \frac{4}{20})+(\frac{7}{21} * \frac{6}{20})+(\frac{9}{21} * \frac{8}{20})$

$P(Same) = \frac{20}{420}+\frac{42}{420} +\frac{72}{420}$

$P(Same) = \frac{20+42+72}{420}$

$P(Same) = \frac{134}{420}$

$P(Same) = \frac{67}{210}$

6 0

2 years ago