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Vera_Pavlovna [14]
3 years ago
7

I NEED HELP ASAP!!!!

Mathematics
1 answer:
N76 [4]3 years ago
8 0
1.5h is the answer you're looking for
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OKAY THIS IS THREE QUESTIONS SO HELP PLS!
blondinia [14]
First one is c
second one is b
and third one is c
4 0
3 years ago
What is the missing number in the table?
Wewaii [24]

Answer:

The missing number is 16.

Step-by-step explanation:

The ratio between x and y is 1:4. 4 times 4 is 16. Hope this helped! :)

4 0
3 years ago
Use the sample information 11formula13.mml = 37, σ = 5, n = 15 to calculate the following confidence intervals for μ assuming th
bija089 [108]

Answer & Step-by-step explanation:

The confidence interval formula is:

I (1-alpha) (μ)= mean+- [(Z(alpha/2))* σ/sqrt(n)]

alpha= is the proposition of the distribution tails that are outside the confidence interval. In this case, 10% because 100-90%

σ= standard deviation. In this case 5

mean= 37

n= number of observations. In this case, 15

(a)

Z(alpha/2)= is the critical value of the standardized normal distribution. The critical valu for z(5%) is 1.645

Then, the confidence interval (90%):

I 90%(μ)= 37+- [1.645*(5/sqrt(15))]

I 90%(μ)= 37+- [2.1236]

I 90%(μ)= [37-2.1236;37+2.1236]

I 90%(μ)= [34.8764;39.1236]

(b)

Z(alpha/2)= Z(2.5%)= 1.96

Then, the confidence interval (90%):

I 95%(μ)= 37+- [1.96*(5/sqrt(15)) ]

I 95%(μ)= 37+- [2.5303]

I 95%(μ)= [37-2.5303;37+2.5303]

I 95%(μ)= [34.4697;39.5203]

(c)

Z(alpha/2)= Z(0.5%)= 2.5758

Then, the confidence interval (90%):

I 99%(μ)= 37+- [2.5758*(5/sqrt(15))

I 99%(μ)= 37+- [3.3253]

I 99%(μ)= [37-3.3253;37+3.3253]

I 99%(μ)= [33.6747;39.3253]

(d)

C. The interval gets wider as the confidence level increases.

8 0
3 years ago
Solve the equation x^2-4x+85=0
lara [203]
Can't factor
use quadratic
for
ax^2+bx+c=0
x=\frac{-b+/- \sqrt{b^2-4ac} }{2a}

given
1x^2-4x+85=0
a=1
b=-4
c=85

x=\frac{-(-4)+/- \sqrt{(-4)^2-4(1)(85)} }{2(1)}
x=\frac{4+/- \sqrt{16-340} }{2}
x=\frac{4+/- \sqrt{-324} }{2}
x=\frac{4+/- (\sqrt{-1})(\sqrt{324}) }{2}
remember that √-1=i
x=\frac{4+/- i(\sqrt{324}) }{2}
x=\frac{4+/- i(18) }{2}
x=2+/-9i

x=2+9i or x=2-9i
3 0
3 years ago
An inverted pyramid is being filled with water at a constant rate of 25 cubic centimeters per second. The pyramid, at the top, h
UNO [17]

Answer:

\frac{225}{16} cm/s

Step-by-step explanation:

We are given that

\frac{dV}{dt}=25cm^3/s

Side of base=4 cm

l=w=4 cm

Height,h=12 cm

We have to find the rate at which the water level rising when the water level is 4 cm.

Volume of pyramid=\frac{1}{3}lwh=\frac{1}{3}l^2h

\frac{l}{h}=\frac{4}{12}=\frac{1}{3}

l=\frac{1}{3}h

Substitute the value

V=\frac{1}{27}h^3

Differentiate w.r.t t

\frac{dV}{dt}=\frac{3}{27}h^2\frac{dh}{dt}

Substitute the values

25=\frac{1}{9}(4^2)\frac{dh}{dt}

\frac{dh}{dt}=\frac{25\times 9}{16}=\frac{225}{16} cm/s

5 0
3 years ago
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