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3 years ago

Frank deposits his savings at the bank. After 3 years, it has earned $102 in interest, at a simple

Mathematics

1 answer:

Liula [17]3 years ago

5 0

Answer:

1530

Step-by-step explanation:

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I have a triangle with angles measuring 40, 40, and 100. Is this one a unique triangle, many triangles, or no triangle?

juin [17]

Answer:

Unique triangle.

Step-by-step explanation:

If you were to draw the triangle, you would've been able to see that one angle is open and that the other 2 were closed.

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3 years ago

A medicine says to drink 1 ounce per 50 pounds of bodyweight per day. If a 150-pound man takes the full dosage for 8 days straig

Korvikt [17]

Answer:

24 ounces

Step-by-step explanation:

Hope this helps.

6 0

3 years ago

The width of a rectangle is 9 inches Shorter than it's length.

m_a_m_a [10]

Divide 252 by 21.
21 times 12 = 252
12 is 9 units shorter than 21.
so the answer is 12 inches by 21 inches
you work backwards from the area
area = width times length
a = wl
we know w is 9 units shorter than the length
so now we have the equation 252=(l-9) times l
plug in 21 for l and then solve there is your answer.

6 0

3 years ago

Read 2 more answers

Write the decimal numeral for ten and five hundredths

mixas84 [53]

10 and 5/100. 5/100 is equal to 0.05, by the definition of decimals, so this turns the expression into 10.05

3 0

2 years ago

Read 2 more answers

67% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 48 owned dogs are r

oksian1 [2.3K]

Answer:

a) Probability that exactly 29 of them are spayed or neutered = 0.074

b) Probability that at most 33 of them are spayed or neutered = 0.66

c) Probability that at least 30 of them are spayed or neutered = 0.79

d) Probability that between 28 and 33 (including 28 and 33) of them are spayed or neutered = 0.574

Step-by-step explanation:

This is a binomial distribution question

probability of having a spayed or neutered dog, p = 0.67

probability of having a dog that is not spayed or neutered, q = 1 - 0.67

q = 0.23

sample size, n = 48

According to binomial distribution formula:

$P(X=r) = nCr p^r q^{n-r}$

where $nCr = \frac{n!}{(n-r)! r!}$

a) Probability that exactly 29 of them are spayed or neutered

$P(X= 29) = 48C29 * 0.67^{29} * 0.23^{19}\\P(X=29) = 0.074$

b) Probability that at most 33 of them are spayed or neutered

$P(X \leq 33) =1 - P(X > 33)\\P(X \leq 33) =1 - 0.34\\P(X \leq 33) = 0.66$

c) Probability that at least 30 of them are spayed or neutered

$P(X \geq 30) = 1 - P(x < 30)\\P(X \geq 30) = 1 - 0.21\\P(X \geq 30) = 0.79$

d) Probability that between 28 and 33 (including 28 and 33) of them are spayed or neutered.

$P(28 \leq X \leq 33) = P(X=28) + P(X=29) + P(X=30) + P(X=31) + P(X=32) + P(X=33)\\P(28 \leq X \leq 33) = 0.053 + 0.074 + 0.095 + 0.112 + 0.121 + 0.119\\P(28 \leq X \leq 33) = 0.574$

8 0

3 years ago