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zmey [24]
2 years ago
10

Can someone help me solve these or explain it to me?

Mathematics
2 answers:
Maurinko [17]2 years ago
6 0
The pic is v blur but i think you can scan google. hope this helped :)
Cloud [144]2 years ago
4 0
The pic is blurred but you can scan it on google ,, hope that helped
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If UV = 6 and TV = 12, what is TU?
love history [14]

Answer:

TU = 6

Step-by-step explanation:

Using the Segment Addition Postulate, we know that TU + UV = TV, and since UV = 6 and TV = 12, we know that TU + 6 = 12, therefore, TU = 6.

3 0
3 years ago
Evaluate log 2 over 3
svetoff [14.1K]
Log 2 over 3 = 0.10034333188
7 0
3 years ago
Southside is purchasing new materials for the school. The school has $1600 to spend on desks and a vending machine. The desks co
Mademuasel [1]

Answer:

15d + 450 \leq 1600

5 0
2 years ago
A monomial of the 2nd degree with a leading coefficient of 3
Rzqust [24]

Given:

Polynomials

To find:

Monomial of 2nd degree with leading coefficient 3

Solution:

Monomial is an algebraic expression with only one term.

Option A: 3 n^{2}-1

It is not a monomial because it have 2 terms.

It is not true.

Option B: 3 n-n^{2}

It is not a monomial because it have 2 terms.

It is not true.

Option C: 3 n^{2}

It have one term only. So, it is a monomial.

Degree means highest power. So degree = 2

Leading coefficient means the value before variable.

Leading coefficient = 3

It is true.

Option D: 2n^3

It have one term only. So, it is a monomial.

Degree means highest power. So degree = 3

It is not true.

Therefore 3 n^{2} is a monomial of 2nd degree with a leading coefficient of 3.

5 0
3 years ago
A real estate agent has 19 properties that she shows. She feels that there is a 30% chance of selling any one property during a
netineya [11]

Answer:

P(X \geq 5)=1-P(X

We can find the individual probabilities:

P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114

P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092

P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358

P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869

P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491

And replacing we got:

P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=19, p=0.3)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

And we want to find this probability:

P(X \geq 5)

And we can use the complement rule:

P(X \geq 5)=1-P(X

We can find the individual probabilities:

P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114

P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092

P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358

P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869

P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491

And replacing we got:

P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178

4 0
3 years ago
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