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3 years ago

From the slope (m) and the y-intercept (b), write the equation of the line : m = 2; b = 3

Mathematics

1 answer:

Anika [276]3 years ago

6 0

Answer:

y= 2x + 3

Step-by-step explanation:

y = mx +b

Plugin the m and the b we have

y=2x + 3

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Factor the gcf out of the polynomial 9x^6+6x^4+15x^3

san4es73 [151]

Answer:3x^3(3x^3+2x+5)

Step-by-step explanation:by looking at the polynomial, you can see that all three shares the variable x. Since 9,6, and 25 all share the GCF of 3, you can factor it out. Since the smallest x value of the polynomial is x^3, that is the x value you want to factor out. Since if you factor out x^3 then 15x^3 would not have a variable anymore.

4 0

4 years ago

a race car is driving under the caution flag at 440 feet per second begins to accelerate at a constant rate after the warning fl

slava [35]

Answer: $\frac{1}{3}\ s$

Step-by-step explanation:

Given

Distance traveled is $s=30t^2+440t$

for s=150 ft

$\Rightarrow 30t^2+440t=150$

$\Rightarrow 30t^2+440t-150=0\\\Rightarrow 3t^2+44t-15=0\\\Rightarrow (3t-1)(t+15)=0\\\Rightarrow t=\dfrac{1}{3}\ s$

6 0

3 years ago

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

bulgar [2K]

Answer:

a) 20.61%

b) 21.82%

c) 42.36%

d) 4 withdrawals

Step-by-step explanation:

This situation can be modeled with a binomial distribution, where p = probability of “success” (completing the course) equals 80% = 0.8 and the probability of “failure” (withdrawing) equals 0.2.

So, the probability of exactly k withdrawals in 20 cases is given by

$\large P(20;k)=\binom{20}{k}(0.2)^k(0.8)^{20-k}$

We are looking for

P(0;20)+P(0;1)+P(0;2) =

$\large \binom{20}{0}(0.2)^0(0.8)^{20}+\binom{20}{1}(0.2)^1(0.8)^{19}+\binom{20}{2}(0.2)^2(0.8)^{18}=$

0.0115292150460685 + 0.0576460752303424 + 0.136909428672063 = 0.206084718948474≅ 0.2061 or 20.61%

Here we want P(20;4)

$\large P(20;4)=\binom{20}{4}(0.2)^4(0.8)^{16}=0.218199402\approx 0.2182=21.82\%$

Here we need

$\large \sum_{k=4}^{20}P(20;k)=1-\sum_{k=1}^{3}P(20;k)$

But we already have P(0;20)+P(0;1)+P(0;2) =0.2061 and

$\large \sum_{k=1}^{3}P(20;k)=0.2061+P(20;3)=0.2061+0.205364 \approx 0.4236=42.36\%$

For a binomial distribution the <em>expectance </em>of “succeses” in n trials is np where p is the probability of “succes”, and the expectance of “failures” is nq, so the expectance for withdrawals in 20 students is 20*0.2 = <em>4 withdrawals.</em>

3 0

3 years ago

Which of the following values are in the range of the function graphed below?

Dafna1 [17]

Answer:

A. 1

Step-by-step explanation:

I see only one red function line in the graph. this line represents the simple function y = 1, for x in the interval [-2,1].

the domain of a function defines the possible values for x, and the range of a function defined the possible values for y.

this function has only one possible value for y : 1

so, only A applies.

5 0

3 years ago

Correct answers only please!

adell [148]

The answer will be 85

5 0

4 years ago

Read 2 more answers