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3 years ago

Find the missing term.

2 answers:

6 0

Find the missing term.7-147 × 798 = 718 × 7-38 × M
7^-147 * 7^98 = 7^18 * 7^-38 * M
7^(-147+98) = 7^(18 + -38)
7^-49 = 7^-20 * M
M = <span>7^-49 / 7^-20
M = 7^(-49 + 20)
M = 7^-29

So the correct answer is </span>C.7^(-29)<span> </span>

dsp733 years ago

5 0

Answer: Option 'C' is correct.

The missing term is $7^{-29}=x$

Step-by-step explanation:

Since we have given that

$7^{-147}\times 7^{98}=7^{18}\times 7^{-38}\times x$

We need to find the missing part i.e. x.

We will use "Exponential law " :

$a^m\times a^n=a^{m+n}$

So, it becomes,

$7^{-147}\times 7^{98}=7^{18}\times 7^{-38}\times x\\\\7^{-147+98}=7^{18-38}\times x\\\\7^{-49}=7^{-20}\times x\\\\7^{-49}\div 7^{-20}=x\\\\7^{-49-(-20)}=x\\\\7^{-49+20}=x\\\\7^{-29}=x$

Hence, the missing term is $7^{-29}=x$

Therefore, Option 'C' is correct.

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Read 2 more answers

The net weight in pounds of a packaged chemical herbicide is uniform for 49.62 < x < 50.04 pounds.

k0ka [10]

Answer:

a) $E(X)=\frac{50.64+49.62}{2}=50.13 pounds$

b) $Var(X)= \frac{(b-a)^2}{12} =\frac{(50.04-49.62)^2}{12}=0.0147$

c) $P(X$

But we see that the distribution is defined ust between 49.62 and 50.04

And in order to find this we can use the CDF (Cumulative distribution function) given by:

$F(X) =\frac{x-a}{b-a}$

And if we replace we got:

$P(X$

And makes sense since all the values are between 49.62 and 50.04

Step-by-step explanation:

For this case we define the random variable X ="net weigth in pounds of a packaged chemical herbicide" and the distribution for X is given by:

$X \sim U(a= 49.62, b=50.04)$

Part a

For the uniform distribution the expected value is given by $E(X) =\frac{a+b}{2}$ where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:

$E(X)=\frac{50.64+49.62}{2}=50.13 pounds$

Part b

The variance for an uniform distribution is given by:

$Var(X)= \frac{(b-a)^2}{12}$

And if we replace we got:

$Var(X)= \frac{(b-a)^2}{12} =\frac{(50.04-49.62)^2}{12}=0.0147$

Part c

For this case we want to find this probability:

$P(X$

But we see that the distribution is defined ust between 49.62 and 50.04

And in order to find this we can use the CDF (Cumulative distribution function) given by:

$F(X) =\frac{x-a}{b-a}$

And if we replace we got:

$P(X$

And makes sense since all the values are between 49.62 and 50.04

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4 years ago

10x^2=6x how do i solve ? quadratic equation

lesantik [10]

Answer:

x=0 or x=6/10

Step-by-step explanation:

10x^2=6x ⇔ 10x^2 - 6x = 0 ⇔ x(10x-6)=0 ⇔ x=0 or x= 6/10

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