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Mamont248 [21]
4 years ago
5

How many 5-millileter test tubes filled with water will it take to fill a 1 leter container?

Mathematics
1 answer:
aliya0001 [1]4 years ago
5 0
One liter is equal to 1000 milliliters. Use the conversion to find your answer. 1000 milliliters / 5mL per tube = 200 tubes
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The polynomial p(x)=x^3+7x^2-36p(x)=x 3 +7x 2 −36p, left parenthesis, x, right parenthesis, equals, x, cubed, plus, 7, x, square
Iteru [2.4K]

Answer:

(x-2)(x+3)(x+6)

Step-by-step explanation:

Given the polynomial function p(x)=x^3+7x^2-36

We are to write it as a product of its linear factor

Assuming the value of x that will make the polynomial p(x) to be zero

Let x = 2

P(2) = 2³+7(2)²-36

P(2) = 8+7(4)-36

P(2) = 8+28-36

P(2) = 0

Since p(2) = 0 hence x-2 is one of the linear factors

Also assume x = -3

P(-3) = (-3)³+7(-3)²-36

P(-3) = -27+7(9)-36

P(-3) = -27+63-36

P(-3) = 36-36

P(-3) = 0

Since p(-3) = 0, hence x+3 is also a factor

The two linear pair are (x-2)(x+3)

(x-2)(x+3) = x²+3x-2x-6

(x-2)(x+3) = x²+x-6

To get the third linear function, we will divide x^3+7x^2-36 by x²+x-6 as shown in the attachment.

x^3+7x^2-36/x²+x-6 = x+6

Hence the third linear factor is x+6

x^3+7x^2-36 = (x-2)(x+3)(x+6)

8 0
3 years ago
Read 2 more answers
Level 4
nignag [31]

132/4 = 33, x = 33 boxes

6 0
3 years ago
Read 2 more answers
I really need help with this
aniked [119]

Answer:

5√202

Step-by-step explanation:

By Pythagoras theorem,

x^2 = 45^2 + 55^2

= 2025 + 3025

x^2 = 5050

x = 5√202

5 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
What is the solution to 4 x + 6 less-than-or-equal-to 18?
Dimas [21]
Im pretty sure its x less-than-or-equal-to 3
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3 years ago
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