All categories

3 years ago

Can someone please help me?

Mathematics

1 answer:

harina [27]3 years ago

4 0

Answer:

F. The mode of the data for Store 2 is greater than the mode of the data for Store 1

Step-by-step explanation:

The mode for Store 2 = 2 (2 appears most)

The mode for Store 1 = 1 (1 appears most)

Therefore, mode of the data for Store 2 is greater than mode of the data for Store 1

Statement F is supported by the information in the dot plots.

You might be interested in

What is 34 ÷ 18 ? Explain your reasoning.

enyata [817]

1.8 that’s my reason

4 0

3 years ago

Read 2 more answers

I don't understand this!

Novosadov [1.4K]

Count the total number of objects. That is the denominator, which goes under the line. Then, only count the stars. That number is the numerator, which goes on top.

4 0

3 years ago

X/w = z/y². solve for y

jek_recluse [69]

Answer:

Step-by-step explanation:

We can simplify (cross multiply) to get $xy^2=wz$

Then we easily simplify to get $y=\sqrt{\frac{wz}{x}},\:y=-\sqrt{\frac{wz}{x}};\quad \:w\ne \:0,\:x\ne \:0$

8 0

3 years ago

Can you plz smile for me?

klio [65]

Answer:

°u°

Step-by-step Explanation

3 0

3 years ago

Read 2 more answers

Example 2: Combining Use of the Multiplication and Addition Rules

bulgar [2K]

Answer:

The probability of getting 6's on both cubes is $\frac{1}{36}$ .

The probability that the total score is at least 11 is $\frac{1}{12}$ .

Step-by-step explanation:

Consider the provided information.

A red cube has faces labeled 1 through 6, and a blue cube has faces labeled in the same way.

Part (A) Both cubes show 6’s.

Probability of getting 6 on red cube is $\frac{1}{6}$

Probability of getting 6 on blue cube is $\frac{1}{6}$

Thus, the probability of getting 6's on both cubes is:

$P(\text{Both 6's})=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$

Hence, the probability of getting 6's on both cubes is $\frac{1}{36}$ .

Part (B) The total score is at least 11.

The possible number of outcomes in which total score is at least 11 is:

Red shows 6 and Blue shows 5.

Blue shows 6 and Red shows 5.

Blue shows 6 and Red shows 6.

Thus, the probability of total score is at least 11.

$P(\text{Total is at least 11})=\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}\\P(\text{Total is at least 11})=\frac{1}{12}$

Hence, the probability that the total score is at least 11 is $\frac{1}{12}$ .

3 0

3 years ago