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Delicious77 [7]
3 years ago
7

A fair number cube with numbers 1-6 was rolled 500 times. The number 4 was landed on 62 times. If the die was rolled 3,000 times

, how many times would it be expected to land on 4, based on the experiment?
a 10
b 372
c 62
d 500
Mathematics
2 answers:
kkurt [141]3 years ago
8 0
The correct answer is B and I hope y’all have a nice day. :3
sweet [91]3 years ago
4 0

Step-by-step explanation:

62/500 is the proportion of rolling 4 to all the rolls, so we just need to expand the fraction so that the denominator is equal to 3000

3000÷500=6

6*62=372 B

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What are the solutions of the equation 6x2 +5x+1 = 0 ?
STatiana [176]

Answer:

Step-by-step explanation:

6x^2+5x+1=0

Descr= b^2-4ac

Descr= 25-24=1

X1= (-b+√descr)/2a = (-5+1)/12= -1/3

X2= (-b-√descr)/2a = (-5-1)/12= -1/2

6 0
3 years ago
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Consider a series circuit consisting of a resistor of R ohms, an inductor of L henries, and variable voltage source of V(t) volt
ser-zykov [4K]

Answer:

I(t)=\frac{1}{3}(1-e^{-30t})

Step-by-step explanation:

We are given that

\frac{dI}{dt}+\frac{R}{L}I=\frac{V(t)}{L}

R=150 ohm

L=5 H

V(t)=10 V

P=\frac{R}{L}=\frac{150}{5}=30

I.F=e^{\int Pdt}=e^{\int 30 dt}=e^{30 t}

I(t)\times I.F=\int e^{30 t}\times 10 dt+C

I(t)\times e^{30 t}=\frac{10}{30}e^{30 t}+C

I(t)=\frac{1}{3}+Ce^{-30 t}

I(0)=0

Substitute t=0

0=\frac{1}{3}+C

C=-\frac{1}{3}

Substitute the values

I(t)=\frac{1}{3}-\frac{1}{3}e^{-30 t}

I(t)=\frac{1}{3}(1-e^{-30t})

5 0
3 years ago
There are 16 tablespoons in one cup. Which table correctly relates the number of cups to the number of tablespoons?
lana [24]

Answer:  The first table

Step-by-step explanation:

1:16

2:32

4:64

8:128

5 0
3 years ago
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Jayce looks around at an assembly. He notices his younger sister to his right and his older brother 48 feet ahead of him. If Jay
olganol [36]

Answer:

I think it's 50

Step-by-step explanation:

48

49

50

51

52

50 8s inbetween the two

6 0
3 years ago
3) g(x)= x3 + x<br> h(x) = x + 4<br> Find g(0)h(0)
Arte-miy333 [17]

Answer:

goh(x)=x^3+12x^2+49x+68

Step-by-step explanation:

g(x) = x^3+x

h(x)=x+4

We have to find

goh(x)

goh(x) = g(h(x))=g(x+4)

If g(x) = x^3+x

g(x+4) = (x+4)^3+(x+4)

           =x^3+12x^2+49x+68

Hence our answer is

goh(x)=x^3+12x^2+49x+68

4 0
3 years ago
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