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dimaraw [331]
3 years ago
14

Which of the following has a quotient of 5?

Mathematics
1 answer:
Vikentia [17]3 years ago
4 0

Answer:

A

Step-by-step explanation:

75÷ 15=5

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Find the value of x that makes the equation true. x + 4 = 12
kirill115 [55]

x + 4 = 12         Subtract 4 on both sides to get "x" by itself

x + 4 - 4 = 12 - 4

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7 0
3 years ago
Hal and Renee play the following game a bag has 11 tiles in it each with a letter from the word probably in it Hal and Renee tak
Masteriza [31]

Answer:

  • <em>D. The game is not fair because the probability of Hal drawing a winning letter is less then the probability of Renee drawing a winning letter </em>

Explanation:

The word is not probably, which has 8 letters, but probability, which has 11 letters

<u>1. Hal</u>

Probability of drawing a vowel

  • The vowels are:  o, a, i, i, and y: 5

  • The total number of letters in the bag is 11.

  • Probability of vowel = 5 / 11

<u>2. Renee</u>

Probability of drawing a consonant

  • The consonants are: p, r, b, b, l, t, and y: 7

  • Probability of a drawing a consonant: 7 /11

<h2>Conclusion</h2>

As you see, the probability of drawing a vowel (5/11), which is the winning letter for Hal, is less than the probability of drawing a consonant (7/11), which is the winning letter of Renee. Then,

  • The game is not fair because the probability of Hal drawing a winning letter is less then the probability of Renee drawing a winning letter.

8 0
4 years ago
Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 10 Allen's hummingb
ruslelena [56]

Answer: provided in the explanation segment

Step-by-step explanation:

(a). from the question, we can see that since that б is known, we can use standard normal, z.

we are asked to find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error?

⇒ 80% confidence interval for the average weight of Allen's hummingbirds is given thus;

x ± z * б / √m

which is

3.15 ± 1.28 * 0.32/√10

= 3.15 ± 0.1295 = 3.0205 or 3.2795

(b). normal distribution of weight (c) б is known

(c). option (a) and (e) are correct

(d).  from the question, let sample size be given as S

this gives';

1.28 * 0.32/√S = 0.15

√S = (1.28 * 0.32) / 0.15 = 2.73

S = 7.4529

cheers i hope this helps

6 0
3 years ago
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