Suppose you spin the spinner once. Find the probability.
A circle is divided equally into eight sections.· Two of the sections are labeled with an upper B.
· One of the sections is labeled with an upper G.
· One of the sections is labeled with an upper Y.
· Four of the sections are labeled with an upper R.
· An arrow originating from the center of the circle is pointing at one of the sections with an upper R.
P(yellow)
A. one-eighth
B. start fraction 1 over 6 end fraction
C. one-fourth
D. one-half.
The probability of landing on yellow is 1/8.
the answer its in the options
Yes, the answer is A. one-eighth.
Suppose you spin the spinner once. Find the probability.
A circle is divided equally into eight sections.· Two of the sections are labeled with an upper B.
· One of the sections is labeled with an upper G.
· One of the sections is labeled with an upper Y.
· Four of the sections are labeled with an upper R.
· An arrow originating from the center of the circle is pointing at one of the sections with an upper R.
P(red or blue)
A. 0
B. one-fourth
C. one-half
D. start fraction 3 over 4 end fraction
The probability of landing on red or blue is 2/8 or 1/4 (since there are 2 red and 2 blue sections out of 8 total sections). Thus, the answer is B. one-fourth.
Drink Survey
Drink
Number of Shoppers Who Preferred It
A
10
B
15
C
7
D
3
E
6
What is the probability that 1 shopper, selected at random, preferred neither Drink E nor Drink C?
A. thirty-five over forty-one
B. thirteen over forty-one
C. twenty-eight over forty-one
D. thirteen over twenty-eight
The number of shoppers who preferred either Drink E or Drink C is 7+6 = 13. Therefore, the number of shoppers who preferred neither is 10+15+3 = 28.
The probability that one shopper, selected at random, preferred neither Drink E nor Drink C is the number of shoppers who preferred neither divided by the total number of shoppers:
P(neither E nor C) = 28/(10+15+7+3+6) = 28/41
So the answer is A. thirty-five over forty-one is not a simplified fraction and is not equal to the probability in this case.
A farmer examines a sample of 25 cartons of eggs and finds that 3 contain cracked eggs. What is the best prediction of the number of cartons with cracked eggs in a delivery with 500 cartons?
A. 6
B. 12
C. 60
D. 120
We can use proportional reasoning to make a prediction. In the sample of 25 cartons, 3 contained cracked eggs. That's a proportion of 3/25 or 0.12.
To predict the number of cartons with cracked eggs in a delivery with 500 cartons, we can multiply this proportion by 500:
0.12 x 500 = 60
So the best prediction of the number of cartons with cracked eggs is 60, or option C.
A coin is tossed. If heads appears, a spinner that can land on any number from 1 to 4 is spun. If tails appears, a second coin is tossed instead of spinning the spinner. What are the possible outcomes?
A. H1 H2 H3 H4
B. H1 H2 H3
C. H1 H2 H3 H4 TH TT
D. HH HT
The possible outcomes are:
- H1
- H2
- H3
- H4
- TH (tails on first coin, heads on second coin)
- TT (tails on both coins)
So the answer is option C, H1 H2 H3 H4 TH TT.
A lunch menu has 4 different sandwiches, 2 different soups, and 5 different drinks. How many different lunches consisting of a sandwich, a soup, and a drink can you choose?
A. 10
B. 11
C. 40
D. 13
To count the number of different lunches we can choose, we multiply the number of options for each part of the lunch together:
4 sandwiches × 2 soups × 5 drinks = 40 different lunches
So the answer is C. 40.
If the spinner is spun twice, what is the probability that the spinner will stop on a consonant and then again on a consonant?
The spinner is a circle divided into 6 equal sections. The sections are labeled L U Z O E and I.
A. two-ninths
B. start fraction 1 over 3 end fraction
C. start fraction 1 over 6 end fraction
D. one-ninth
7 / 13
There are four consonants on the spinner: L, Z, O, and N (we treat "Y" as a vowel here). There are a total of 6 sections, so the probability of stopping on a consonant on the first spin is 4/6 or 2/3.
After the first spin, there are only 5 sections left and 3 of them are consonants (since we're treating "Y" as a vowel), so the probability of stopping on a consonant on the second spin, given that we stopped on a consonant on the first spin, is 3/5.
To find the probability of both events occurring (stopping on a consonant, then stopping on a consonant again), we multiply their probabilities:
2/3 × 3/5 = 6/15 = 2/5
So the answer is not among the given options, it is 2/5.