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notsponge [240]
3 years ago
10

A student solved the equation 2x+2=10 using algebra tiles. She incorrectly says the solution is 6 . Solve the equation. What mis

take might the student have​ made?

Mathematics
2 answers:
Alex_Xolod [135]3 years ago
7 0

Answer:

she added 2 instead of subtracting it. the answer is x=4

Step-by-step explanation:

10+2=12

12/2=6

enot [183]3 years ago
3 0

Answer:

Step-by-step explanation:

2x+2=10 subtract 2 from each side

2x=8 divide each side by 2

x=4 is the correct solution.

Her error was most likely subtracting 2 from the left and adding 2 to the right

2x+2=10

2x=12

x=6

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It helps you see the equation in "picture" form.

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4 0
3 years ago
Someone please help i do not understand this
Andrej [43]

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3 years ago
Julie is digging a square area in her backyard a for a vegtable garden. If the area is 52 square feet, what is the approximate l
AURORKA [14]

Answer:

<em>The correct option will be:   (b) 7 feet.</em>

Step-by-step explanation:

Suppose, the length of the garden is  x feet.

As the <u>shape of the garden is like a square, then the area of the garden</u> will be: (length)^2 = x^2 feet²

Given that, the <u>area is 52 square feet</u>. So, the equation will be.......

x^2= 52\\ \\ x=\sqrt{52} = 7.211.... \approx 7

So, the approximate length of the garden will be 7 feet.

4 0
3 years ago
The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont
BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)

and its cumulative distribution function (CDF) is

\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so  A∩B = B and  

\bf P(A | B) = P(B)P(B) = (P(B))^2

We have  

\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237

hence,

\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}

<em>Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection. </em>

This means that,

<em>Probability that at least one of them needs replacement at the first inspection =  </em>

\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021

5 0
3 years ago
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