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4 years ago

Please answer!! Will mark brainliest!!!

Mathematics

1 answer:

ddd [48]4 years ago

8 0

Answer:

C & D

Step-by-step explanation:

$5^{12} * 5^{8} = 5^{12+8} = 5^{20}\\\\(5^{3}*5^{2})^{4} = (5^{3+2})^{4} =(5^{5})^{4} = 5^{5*4}=5^{20}\\\\(5^{5})^{4}=5^{5*4}=5^{20}$

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(x + 4)2 – 3(x + 4) – 3 = 0 Select the solution(s) of the original equation.

Aneli [31]

Answer:

x=−7

Step-by-step explanation:

(x+4)(2)−3(x+4)−3=0

(x)(2)+(4)(2)+(−3)(x)+(−3)(4)+−3=0(Distribute)

2x+8+−3x+−12+−3=0

(2x+−3x)+(8+−12+−3)=0(Combine Like Terms)

−x+−7=0

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3 years ago

The____ function is the most basic function in a family of functions

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The answer is: parent function

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4 years ago

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The top graph represents f(x)

Flura [38]

Answer:

Graph number 1 should be the answer

Step-by-step explanation:

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2 years ago

Given: FG is an altitude. What can you conclude?<br><br> O

sesenic [268]

Answer:

A or B

Step-by-step explanation:

Its more likely B

3 0

3 years ago

Gears produced by a grinding process are categorized either as conforming (suitable for their intended purpose), downgraded (uns

Svetach [21]

Answer:

A. P=0.4013

B. P=0.9884

C. P=0.3222

Step-by-step explanation:

We can model this with a binomial random variable.

The sample size is n=10.

The probability is p=0.05 for scrap and p=0.05+0.15=0.20 for degraded or scrap.

The probability of having k scrap gears in the sample is:

$P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{10}{k} 0.05^{k} 0.95^{10-k}\\\\\\$

The probability that one or more is scrap can be calculated as 100% less the probability that no one is scrap:

$P(x\geq1)=1-P(x=0)\\\\\\P(x=0) = \dbinom{10}{0} p^{0}(1-p)^{10}=1*1*0.5987=0.5987\\\\\\P(x\geq1)=1-0.5987=0.4013$

The probability that 8 or more are not scrap is equal to the probability of having 2 or less that are scrap:

$P(x\leq2)=P(x=0)+P(x=1)+P(x=2)\\\\\\P(x=0) = \dbinom{10}{0} p^{0}(1-p)^{10}=1*1*0.5987=0.5987\\\\\\P(x=1) = \dbinom{10}{1} p^{1}(1-p)^{9}=10*0.05*0.6302=0.3151\\\\\\P(x=2) = \dbinom{10}{2} p^{2}(1-p)^{8}=45*0.0025*0.6634=0.0746\\\\\\P(x\leq2)=0.5987+0.3151+0.0746=0.9884$

The probability that more than two are degraded or scrap (p=0.2) is calculated as:

$P(x>2)=1-P(x\leq2)=1-(P(x=0)+P(x=1)+P(x=2))\\\\\\P(x=0) = \dbinom{10}{0} p^{0}(1-p)^{10}=1*1*0.1074=0.1074\\\\\\P(x=1) = \dbinom{10}{1} p^{1}(1-p)^{9}=10*0.2*0.1342=0.2684\\\\\\P(x=2) = \dbinom{10}{2} p^{2}(1-p)^{8}=45*0.04*0.1678=0.3020\\\\\\P(x>2)=1-(0.1074+0.2684+0.3020)=1-0.6778=0.3222$

4 0

3 years ago