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2 years ago

5

Justin is measuring his yard for a fence. He needs to figure out how much wood it will take. What unit of measure does Justin ne

ed to use?

1 answer:

sineoko [7]2 years ago

6 0

Answer:

i think it is miles yay

You might be interested in

I need help with this. I'll mark brainliest to whoever is correct

zaharov [31]

Answer:

As shown in picture, the area is divided into 4 parts: 3 triangles and 1 rectangles.

Total area:

A = smallest triangle + medium triangle + largest triangle + rectangle

= 2 x 2 x 1/2 + 2 x 6 x 1/2 + 4 x (2 + 2 + 6) x 1/2 + 2 x 2

= 2 + 6 + 20 + 4

= 32

Hope this helps!

:)

8 0

2 years ago

A marble is selected at random from a jar containing 3 red marbles, 4 yellow marbles, and 6 green marbles. What is the probabi

alexgriva [62]

Answer: a) $\dfrac{3}{13},\ b)\dfrac{9}{13}\ c)\dfrac{9}{13},\ d)\dfrac{4}{13}$

Step-by-step explanation:

Since we have given that

Number of red marbles = 3

Number of yellow marbles = 4

Number of green marbles = 6

Total number of marbles = 3 +4 +6 = 13

a) the marble is red?

Probability of getting a red marble = $\dfrac{3}{13}$

b) the marble is not yellow?

Probability of not getting yellow = $\dfrac{3+6}{13}=\dfrac{9}{13}$

c) the marble is either red or green?

Probability of getting either red or green = P(red) + P(green) = $\dfrac{3}{13}+\dfrac{6}{13}=\dfrac{9}{13}$

d) the marble is neither red nor green?

Probability of getting neither red nor green = P(getting yellow) = $\dfrac{4}{13}$

Hence, a) $\dfrac{3}{13},\ b)\dfrac{9}{13}\ c)\dfrac{9}{13},\ d)\dfrac{4}{13}$

5 0

2 years ago

A figure is composed of two half-circles.

posledela

Answer:

The answer is C, 220 units.

3 0

3 years ago

Assume that body masses of Goldfinch birds follow a normal distribution with standard deviation equal to 0.04 oz. An ornithologi

il63 [147K]

Answer:

The formula to generate 70% confidence interval is: $[\overline{x} - 0.013, \overline{x} + 0.013]$ , in which $\overline{x}$ is the sample mean.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.7}{2} = 0.15$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.15 = 0.85$ , so Z = 1.037.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Assume that body masses of Goldfinch birds follow a normal distribution with standard deviation equal to 0.04 oz.

This means that $\sigma = 0.04$ .

Sample of 10 birds:

This means that $n = 10$ .

The margin of error is of:

$M = z\frac{\sigma}{\sqrt{n}}$

$M = 1.037\frac{0.04}{\sqrt{10}}$

$M = 0.013$

The lower end of the interval is the sample mean of $\overline{x}$ subtracted by M.

The upper end of the interval is the sample mean of $\overline{x}$ added to M.

Then, the formula to generate 70% confidence interval is: $[\overline{x} - 0.013, \overline{x} + 0.013]$ , in which $\overline{x}$ is the sample mean.

8 0

3 years ago

Kristin decides to spend at most $50 for a birthday dinner at a restaurant, including a 15% tip.,

Dmitry [639]

0.15x + x <(or equal to) 50
Where x is the meal's cost (no tip included)

50 - 15% = 42.5.
x >(or equal to) 42.50.

(nothing is copied content, I copy pasted a number into the right position, feel free to check it mods)

7 0

3 years ago