Classify the following triangle. Check all that apply

Students did push-ups for a fitness test. Their goal was 20 push-ups. For each student,

ale4655 [162]

Answer:

1- 80%

2- 210%

3- 65%

4- 105%

Step-by-step explanation:

5 0

2 years ago

Is 5.692 a terminating or a repeating decimal

Ostrovityanka [42]

5.692 is a terminating decimal because the decimal stopped at the digit of 2.

Hope this helps!

7 0

3 years ago

Read 2 more answers

The Internal Revenue Service estimates that 8% of all taxpayers filling out forms make mistakes. An IRS employee announces at lu

belka [17]

Answer:

Mean = 2

Standard deviation = 1.356

Step-by-step explanation:

We are given;

Probability of taxpayers that make mistakes when filling out forms: p = 0.08

Sample size; n = 25

In binomial distribution;

Means is;

μ = np

Thus;

μ = 25 × 0.08

μ = 2

Formula for standard deviation here is;

σ = √(np(1 - p))

σ = √(25 × 0.08(1 - 0.08))

σ = 1.356

7 0

3 years ago

How many centiliters are in 6.02 decaliters?

Aloiza [94]

1 centiliter is .01 of a liter.

6.02 decaliter is 60.2 liters.

6.02 decaliter is also 6020 centiliters

So 6.02 decaliter is 6020 centiliters

6 0

3 years ago

A certain company sends 40% of its overnight mail parcels by means of express mail service A1. Of these parcels, 4% arrive after

Harrizon [31]

Answer:

(a) The probability that a randomly selected parcel arrived late is 0.026.

(b) The probability that a parcel was late was being shipped through the overnight mail service A₁ is 0.615.

(c) The probability that a parcel was late was being shipped through the overnight mail service A₂ is 0.192.

(d) The probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.

Step-by-step explanation:

Consider the tree diagram below.

(a)

The law of total probability sates that: $P(A)=P(A|B)P(B)+P(A|B')P(B')$

Use the law of total probability to determine the probability of a parcel being late.

$P(L)=P(L|A_{1})P(A_{1})+P(L|A_{2})P(A_{2})+P(L|A_{3})P(A_{3})\\=(0.04\times0.40)+(0.01\times0.50)+(0.05\times0.10)\\=0.026$

Thus, the probability that a randomly selected parcel arrived late is 0.026.

(b)

The conditional probability of an event A provided that another event B has already occurred is:

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

Compute the probability that a parcel was late was being shipped through the overnight mail service A₁ as follows:

$P(A_{1}|L)=\frac{P(L|A_{1})P(A_{1})}{P(L)} \\=\frac{0.04\times 0.40}{0.026} \\=0.615$

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₁ is 0.615.

(c)

Compute the probability that a parcel was late was being shipped through the overnight mail service A₂ as follows:

$P(A_{2}|L)=\frac{P(L|A_{2})P(A_{2})}{P(L)} \\=\frac{0.01\times 0.50}{0.026} \\=0.192$

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₂ is 0.192.

(d)

Compute the probability that a parcel was late was being shipped through the overnight mail service A₂ as follows:

$P(A_{3}|L)=\frac{P(L|A_{3})P(A_{3})}{P(L)} \\=\frac{0.05\times 0.10}{0.026} \\=0.192$

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.

4 0

3 years ago