Ask question

Ask question

All categories

3 years ago

10

Estimate each product.Then write whether the estimate is greater than or less than the actual product.There are 62 rows of 9 cha

irs in the movie theater.About how many chairs are there?

1 answer:

Westkost [7]3 years ago

8 0

That is 62x9 which is 9x2=18 and 9x60=540 so you then add 540+18=558 chairs

You might be interested in

There are 874 pieces of candy in the piñata. There are 23 children at the party, and they all agreed to split the candy equally.

blsea [12.9K]

Answer:

38 candies.

Step-by-step explanation:

Given the following data;

Total number of candy = 874

Total number of children = 23

Since we know that they are all sharing it equally, each child would get;

Each child = 874/23

Each child = 38 candies.

6 0

3 years ago

You placed a 11% down payment on a new home. Given that the amount of the down payment was $18,210.18, determine the price of th

Reil [10]

D its D gotta be 20 charecters

7 0

3 years ago

Read 2 more answers

Helppppppppppppppppppppppppp

Solnce55 [7]

Set the two expressions equal to each other.

7x + 3 = 9x

Simplify, isolate the x, subtract 7x from both sides

7x (-7x) + 3 = 9x (-7x)
3 = 2x

Set the equation = 0, subtract 3 from both sides

3 (-3) = 2x (-3)

0 = 2x - 3

2x - 3, or D, should be your answer

hope this helps

8 0

3 years ago

Max and Jenna are each making pastries for their family reunion. Max is making beignets with a special type of flour. For every

Ipatiy [6.2K]

Jenna can make 48 servings

5 0

3 years ago

The length of human pregnancies from conception to birth follows a distribution with mean 266 days and standard deviation 15 day

Katena32 [7]

Answer:

a) 81.5%

b) 95%

c) 75%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 266 days

Standard Deviation, σ = 15 days

We are given that the distribution of length of human pregnancies is a bell shaped distribution that is a normal distribution.

Formula:

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

a) P(between 236 and 281 days)

$P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{281-266}{15})\\\\= P(-2 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -2)\\= 0.838 - 0.023 = 0.815 = 81.5\%$

b) a) P(last between 236 and 296)

$P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{296-266}{15})\\\\= P(-2 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2)\\= 0.973 - 0.023 = 0.95 = 95\%$

c) If the data is not normally distributed.

Then, according to Chebyshev's theorem, at least $1-\dfrac{1}{k^2}$ data lies within k standard deviation of mean.

For k = 2

$1-\dfrac{1}{(2)^2} = 75\%$

Atleast 75% of data lies within two standard deviation for a non normal data.

Thus, atleast 75% of pregnancies last between 236 and 296 days approximately.

7 0

3 years ago