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3 years ago

What would be the total number of outcomes in the sample space?

Mathematics

2 answers:

snow_tiger [21]3 years ago

8 0

20x26=520

The answer is D 520.

Hope it helps.

shusha [124]3 years ago

4 0

I believe the answer is d 520

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Solve the following system algerbraically: y=-3(x-2) squared+4 y=-6x+16

pishuonlain [190]

$y = - 3(x - 2) ^{2} + 4$
$y = - 6x + 16$
distribute the -3 and the square into the first equation
$y = - 3x ^{2} + 12x - 8$
$y = - 6x + 16$
multiply the second equation by 2
$y = - 3x ^{2} + 12x - 8$
$y = - 12x + 32$
add the equations together
$y = - 3x ^{2} + 24$
factor out -3
$y = - 3(x ^{2} - 8)$
factor inside parenthesis
$y = - 3(x + \sqrt{8} )( x - \sqrt{8})$
refine
$y = - 3(x + 2 \sqrt{2} )(x - 2 \sqrt{2} )$

8 0

3 years ago

In a study, 8 people ate a total of 1,312 pounds of potatoes in 2 years.

cricket20 [7]

Answer:

83.125 pounds of potatoes

Step-by-step explanation:

1. Divide 1,312 by 2

1,312/2 = 665

This means that in 1 year 8 people ate 665 pounds of potatoes.

2. Divide 665 by 8

665/8 = 83.125

This means that in 1 year 1 person ate 83.125 pounds of potatoes.

5 0

3 years ago

|14+x|-5=k solve for k

guajiro [1.7K]

Answer:

k=|14+x|-5

Step-by-step explanation:

just rewrite the equation as k=|14+x|-5

6 0

3 years ago

Find the length of B to the nearest tenth using the Pythagorean theorem.

bekas [8.4K]

Step-by-step explanation:

Here

BC=P=10

AC=B=b

AB=H=15

then using fourmula

h^2= p^2+ b^2

b^2= 15×15 - 10× 10

b^2= 125

b=11.2

4 0

3 years ago

The mean amount purchased by a typical customer at Churchill’s Grocery Store is $23.50, with a standard deviation of $5.00. Assu

m_a_m_a [10]

Answer: 0.0170

Step-by-step explanation:

Given : The mean amount purchased by a typical customer at Churchill’s Grocery Store is $23.50, with a standard deviation of $5.00.

i.e. $\mu=23.50$

$\sigma=5$

We assume the distribution of amounts purchased follows the normal distribution.

Sample size : n=50

Let $\overline{x}$ be the sample mean.

Formula : $z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

Then, the probability that the sample mean is at least $25.00 will be :-

$P(\overline{x}\geq\25.00)=P(\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\geq\dfrac{25-23.50}{\dfrac{5}{\sqrt{50}}})\\\\=P(z\geq2.12)\\\\=1-P(z$

Hence, the likelihood the sample mean is at least $25.00= 0.0170

5 0

4 years ago