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3 years ago

Number of pennies, nickels, dimes, quarters, and other types of currency contained in a piggy bank:

Mathematics

1 answer:

PIT_PIT [208]3 years ago

7 0

The answer is A. because that is a sample answer that doesnt has the complete information

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The first day, Brian saw 31 birds and twice as many squirrels as birds. The second day, Brian saw 20 birds and 42 squirrels. Whi

ladessa [460]

Answer: A good estimation would be about 150.

Step-by-step explanation:

To estimate we can round instead of using exact numbers.

First day he saw about 30 birds and twice as many squirrels, meaning about 60 squirrels. This is about 90 total when added together.

Second day he saw 20 birds and about 40 squirrels so about 60 when added together.

Add day one and day two totals for...

90+60=150

8 0

2 years ago

Cual es el 45% de 18000

cricket20 [7]

8100 es el 45% de 18000

3 0

3 years ago

PLZ HURRY!!!! 15 POINTS!

Daniel [21]

I think it’s 7.5 minutes

5 0

3 years ago

Read 2 more answers

Please help will give brainliest if correct. What’s the equation of line a?

attashe74 [19]

Answer:

y = 14/5x + 2

Step-by-step explanation:

is seems as though the y-intercept would be 2 and the slope is 2.8 which is

2 4/5 or 14/5

therefore, equation would be y = 14/5x + 2

7 0

2 years ago

Read 2 more answers

The port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest

Ksenya-84 [330]

Answer:

a) 0.7287

b) 0.9663

c) 0.237

d) 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.5 million tons of cargo per week

Standard Deviation, σ = 0 .82 million tons

We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P( port handles less than 5 million tons of cargo per week)

P(x < 5)

$P( x < 5) = P( z < \displaystyle\frac{5 - 4.5}{0.82}) = P(z < 0.609)$

Calculation the value from standard normal z table, we have,

$P(x < 5) =0.7287= 72.87\%$

b) P( port handles 3 or more million tons of cargo per week)

$P(x \geq 3) = P(z \geq \displaystyle\frac{3-4.5}{0.82}) = P(z \geq −1.82926)\\\\P( z \geq −1.82926) = 1 - P(z < -1.829)$

Calculating the value from the standard normal table we have,

$1 - 0.0337 = 0.9663 = 96.63\%\\P( x \geq 3) = 96.63\%$

c)P( port handles between 3 million and 4 million tons of cargo per week)

$P(3 \leq x \leq 4) = P(\displaystyle\frac{3 - 4.5}{0.82} \leq z \leq \displaystyle\frac{4-4.5}{0.82}) = P(-1.829 \leq z \leq -0.609)\\\\= P(z \leq -0.609) - P(z < -1.829)\\= 0.271-0.034 = 0.237= 23.7\%$

$P(3 \leq x \leq 4) = 23.7\%$

d) P(X=x) = 0.85

We have to find the value of x such that the probability is 0.85.

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 4.5}{0.82})=0.85$

$= 1 -P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.85$

$=P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.15$

Calculation the value from standard normal z table, we have,

$P( z \leq -1.036) = 0.15$

$\displaystyle\frac{x - 4.5}{0.82} = -1.036\\x = 3.65$

Thus, 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

8 0

2 years ago