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4 years ago

50r − 6 = 82r − 70. a. r=64 b. r= -64 c.r=2 d.r=-2

Mathematics

1 answer:

lisabon 2012 [21]4 years ago

3 0

Answer:

Step-by-step explanation:

Hope this helps.

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2. TOYS A playground ball has a radius of

dalvyx [7]

Answer:

1767.1 inches cubed

Step-by-step explanation:

The volume of a ball, or a sphere, is: $V=\frac{4}{3} \pi r^3$ , where r is the radius.

Here, the radius is r = 7.5. So, plug in this value: $V=\frac{4}{3} \pi *(7.5)^3=562.5\pi$

562.5 pi is about 1767.1 inches cubed.

Thus, the volume is 1767.1 inches cubed.

Hope this helps!

6 0

3 years ago

Read 2 more answers

The base of a triangle prism shown is an isosceles triangle having a height of H cm, and an area of 36cm squared

Alika [10]

Answer:

the value of h on the diagram is 9cm

Step-by-step explanation:

area = 1/2 * base * height

a = 1/2*b*h

36=1/2*8*h

36=4h

since the 1/2 will multiply the 8

36=4h

multiply both sides by 4

h=9cm.

4 0

3 years ago

How many significant digits are in 705.0 ?

ICE Princess25 [194]

All trailing zeros after dec. pt are sig.
all nonzero numbers are sig.
all zeros between nonzero numbers are sig.

there are 4 significant digits

7 0

3 years ago

Read 2 more answers

what is the constant of variation? I determined that it didn't have one.. but that isn't an option? Help please!

lesantik [10]

The constant of a variable is my+b

8 0

4 years ago

The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typo

muminat

Answer:

The required probability is 0.55404.

Step-by-step explanation:

Consider the provided information.

The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typographical errors on a page of second booklet is a Poisson random variable with mean 0.3.

Average error for 7 pages booklet and 5 pages booklet series is:

λ = 0.2×7 + 0.3×5 = 2.9

According to Poisson distribution: $P(k{\text{ events in interval}})={\frac {\lambda ^{k}e^{-\lambda }}{k!}}$

Where $\lambda$ is average number of events.

The probability of more than 2 typographical errors in the two booklets in total is:

$P(k > 2)= 1 - {P(k = 0) + P(k = 1) + P(k = 2)}$

Substitute the respective values in the above formula.

$P(k > 2)= 1 - ({\frac {2.9 ^{0}e^{-2.9}}{0!}} + \frac {2.9 ^{1}e^{-2.9}}{1!}} + \frac {2.9 ^{2}e^{-2.9}}{2!}})$

$P(k > 2)= 1 - (0.44596)$

$P(k > 2)=0.55404$

Hence, the required probability is 0.55404.

4 0

3 years ago