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2 years ago

Alfred tosses a quarter and then rolls a six-sided number cube. What is the total number of outcomes for this event?

Mathematics

1 answer:

algol132 years ago

8 0

Answer:

Step-by-step explanation:

quarter and 1

quarter and 2

quarter and 3

quarter and 4

quarter and 5

quarter and 6

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00:57

prisoha [69]

<h3>Given:</h3>

h= 19 ft
r= 12 ft

h= Height
r= Radius

The volume of the given cone.

<h3>Solution:</h3>

$\large\boxed{Formula:V= \frac{1}{3}\pi{r}^{2}h}$

$\large\boxed{\red \pi \red = \red 3 \red . \red 1 \red 4}$

Let's solve!

Substitute the values according to the formula.

$V=\frac{1}{3}×3.14×{12}^{2}×19$

$\large\boxed{V=2863.68 \: {ft}^{3}}$

Therefore, the volume of the given cone is 2863.68 cubic feets.

5 0

2 years ago

Trudy works at a job centre.

denpristay [2]

To solve the problem, to 40% of the total customer should be solve. And this can be solve by muliplting it by the fraction (0.4). so 140 x 0.4 = 56 customers. Since Trudy has already 60 of her sutomers job ready then she met her target.

3 0

3 years ago

A school has a goal of getting at least 2000 boxtops. They have 872 so far. They have 8 days left to get it. At least how many d

True [87]

Answer:

School needs to collect at least 144 box tops each day to meet their goal.

Step-by-step explanation:

The total requirement of box tops = 2,000

The number of box tops they have already received = 872

So, the box tops left = Total number required- Number of tops received

=2,000 - 872

= 1,128 box tops are left so far.

Number of days left = 8

So, number of minimum tops needed each day = $\frac{\textrm{Total tops required}}{\textrm{Number of days left}}$

= $\frac{1128}{8} = 141$

Hence, they need to collect at least 144 box tops each day to meet their goal.

5 0

2 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

3 years ago

What is the simplified expression for the expression below?

kotykmax [81]

Answer:

Step-by-step explanation:

- 1/5 of 5x is -1 or just -x and -1/5 of 20 is -4. Then the other one is-1/4 of 4 is -x and -1/4 of -28 is 7. Now combine like terms so -x + -x is -2x and 7+ -4 or 7-4 is 3

8 0

2 years ago