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Natali [406]
3 years ago
12

jll invited 37 people to her birthday party they each ate 8 pieces of pizza how many pieces of pizza did they eat

Mathematics
1 answer:
Molodets [167]3 years ago
8 0
If Jill invited 37 people to her party, there will be a total of 38 people there. If each 38 people had 8 pieces of pizza, you would multiply 38 with 8, which the answer will be 304. They ate 304 pieces of pizza.
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Solve for the first variable in one of the equations, then substitute the result into the other equation.
Point Form: (-1,0)
Equation form: x=-1,y=0
4 0
4 years ago
2. In segment AC, the midpoint is B.<br> If segment AC = 2 and AB = 7x-3, find x.
Scorpion4ik [409]
X=1
midpoint is half way, half of 2 is 1
4 0
3 years ago
9 4/5 - 2 3/10 in simplest form
iris [78.8K]

first multiply 4/5 by 2

=8/10

Subtract 8 and 3

=5

Subtract 9 and 2

=7

New fraction is: 7 5/10

Reduce to 7 and 1/2

Answer: 7 1/2

6 0
3 years ago
Find the perimeter of squared garden if its area is 1369m2​
irina [24]

Answer:

the perimeter is of 148 m

Step-by-step explanation:

the perimeter is the sum of all sides so an square has all sides equal so we have

P = x+x+x+x = 4x

and the area is

A=x^2=1369m^2

so we find x

\sqrt{x^2}=\sqrt{1369}\\\\x=37

and now the perimeter

P=4x\\\\P=4(37)\\\\P=148

6 0
3 years ago
Let Y1,Y2, . . . ,Yn be a random sample from a normal distribution where the mean is 2 and the variance is 4. How large must n b
Marina86 [1]

Answer:

n= 60

Step-by-step explanation:

Hello!

You have Y₁, Y₂, ..., Yₙ random sample with a normal distribution: Y~N(μ;σ²)

μ= 2

σ²= 4

You need to calculate a sample size n so that (1.9 ≤ Y ≤2.1)= 0.99

To reach the sample size you need to work with the distribution of the sample mean (Y[bar]) because it is this distribution that is directly affected by the sample size.

Y[bar]~N(μ;σ²/n)

Under the sample mean distribution you have to use the standard normal:

Z=  Y[bar] - μ  ~N(0;1)

σ/√n

Now the asked interval is:

P(1.9 ≤ Y[bar] ≤2.1)= 0.99

The upper bond is 2.1

The lower bond is 1.9

The difference between the two bonds is the amplitude of the interval a=2.1-1.9= 0.2

And the probability included between these two bonds is 0.99

With this in mind you can rewite it as an interval for the sample mean:

Y[bar] + Z_{1-\alpha /2}*(σ/√n) - (Y[bar] + Z_{1-\alpha /2}*(σ/√n))= 0.2

Using the semiamplitude (d) of the interval you can easly calculate the required sample:

d= a/2= 0.2/2= 0.1

d= Z_{1-\alpha /2}*(σ/√n)

d* Z_{1-\alpha /2}= σ/√n

√n*(d* [tex]Z_{1-\alpha /2}[/tex)= σ

√n= σ/(d* [tex]Z_{1-\alpha /2}[/tex)

n= (σ/(d* [tex]Z_{1-\alpha /2}[/tex))²

n= (2/(0.1* 2.586))²

n= 59,81 ≅ 60

I hope it helps!

4 0
3 years ago
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