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3 years ago

I need help on these questions

Mathematics

2 answers:

Naya [18.7K]3 years ago

8 0

I think 7,a is 8 cm :)

victus00 [196]3 years ago

5 0

6a. f = 25.
6b. 1.
7a. 2.
7b. 4.
7c. 2=z.
7d. the circumference for z is 12.56.

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A fruit bowl contained 2 bananas, 5 oranges, and 4 apples. In simplest form, what is the ratio of apples to bananas in the fruit

Lisa [10]

Answer:

1:2

Step-by-step explanation:

there are 4 apples and 2 bananas so you divide 4/2 and get 1:2

8 0

3 years ago

Read 2 more answers

WILL MARK BRAINLIEST!!! GIVING 15 POINTS!

FinnZ [79.3K]

Answer:

#1) Angle measure of B

The angles in a triangle should add up to 180. Since we know the other two angle measures, we can subtract it from 180. (180-45-15= Angle B)

Angle B= 120

#2) Length of segment BC

I would use the Law of Sines. So, sin120/200 = sin45/BC

Then cross multiply and divide.

Segment BC= 163.3

#3) Length of Segment BA

Again, I would use the Law of Sines. Same process, sin120/200 = sin15/BA

Segment BA= 59.8

6 0

4 years ago

Suppose the rate of type II diabetes in 40- to 59-year-olds is 7% among Caucasians, 10% among African-Americans, 12% among Hispa

melomori [17]

Answer:

The overall probability of type II diabetes among 40- to 59-year-olds in Houston is 9.3%.

Step-by-step explanation:

We have these following rates of type II diabetes:

7% among Caucasians

10% among African-Americans

12% among Hispanics

5% among Asian-Americans.

The ethnic distribution of Houston is:

30% Caucasian

25% African-American

40% Hispanic

5% Asian-American

What is the overall probability of type II diabetes among 40- to 59-year-olds in Houston?

$P = P_{1} + P_{2} + P_{3} + P_{4}$

$P_{1}$ is the probability of finding a Caucasian with type II diabetes in Houston. So it is 7% of 30%.

$P_{1} = 0.07*0.30 = 0.021$

$P_{2}$ is the probability of finding an African-American with type II diabetes in Houston. So it is 10% of 25%.

$P_{2} = 0.1*0.25 = 0.0215$

$P_{3}$ is the probability of finding a Hispanic with type II diabetes in Houston. So it is 12% of 40%.

$P_{3} = 0.12*0.40 = 0.048$

$P_{4}$ is the probability of finding an Asian-American with type II diabetes in Houston. So it is 5% of 5%.

$P_{3} = 0.05*0.05 = 0.0025$

$P = P_{1} + P_{2} + P_{3} + P_{4} = 0.021 + 0.0215 + 0.048 + 0.0025 = 0.093$

The overall probability of type II diabetes among 40- to 59-year-olds in Houston is 9.3%.

4 0

4 years ago

An official for a national dog show studies the characteristics of one breed of dog, the Dandie Dinmont Terrier. Two common meas

LUCKY_DIMON [66]

Answerb:

Step-by-step explanation:

8 0

3 years ago

Based on experience, the probability that an incoming email is spam (an undesired message of solicitation) is 38%. If three emai

SCORPION-xisa [38]

Answer:

0.0549

0.7617

0.2383

Step-by-step explanation:

Given that:

Probability that incoming email is spam ;

P(spam) = 0.38

If three emails are in my inbox, what is the probability that:

A. All three messages are spam.

P(1st message being spam) * p(second message being spam) * p(3rd message being spam)

= (0.38) * (0.38) * (0.38)

= 0.054872

= 0.0549

B. At least one of the three messages are spam.

1 - p(none of the 3 messages are spam)

P(none of the 3 messages are spam)

P(not spam) = 1 - p(spam) = 1 - 0.38 = 0.62

1 - [P(not spam) * p(not spam) * p(not spam)]

1 - (0.62 * 0.62 * 0.62)

= 1 - 0.238328

= 0.761672

= 0.7617

C. None of the three messages are spam.

P(not spam) = 1 - p(spam) = 1 - 0.38 = 0.62

P(not spam) * p(not spam) * p(not spam)

(0.62 * 0.62 * 0.62)

= 0.238328

= 0.2383

7 0

3 years ago