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3 years ago

Answer the question below. Type your response in the space provided.

Mathematics

2 answers:

Naya [18.7K]3 years ago

7 0

Answer:54

Step-by-step explanation:

(12 x 4) = 48

(12 x .5) = 6

48 + 6 = 54

Anna35 [415]3 years ago

6 0

Answer:

(12 x 4) = 48

(12 x .5) = 6

48 + 6 = 54

the correct answer is 54

plz mark me as brainliest if this helped :)

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Write the sum as a product of the GCF and a sum: 39 + 91

zysi [14]

13(3 + 7)

Hope I Helped You!!! :-)

Have A Good Day!!!

4 0

3 years ago

An Internet reaction time test asks subjects to click their mouse button as soon as a light flashes on the screen. The light is

Ganezh [65]

Answer:

(a) 0.20

(b) 31%

Step-by-step explanation:

The random variable Y models the amount of time the subject has to wait for the light to flash.

The density curve represents that of an Uniform distribution with parameters a = 1 and b = 5.

So, $Y\sim Unif(1,5)$

(a)

The area under the density curve is always 1.

The length is 5 units.

Compute the height as follows:

$\text{Area under the density curve}=\text{length}\times \text{height}$

$1=5\times\text{height}\\\\\text{height}=\frac{1}{5}\\\\\text{height}=0.20$

Thus, the height of the density curve is 0.20.

(b)

Compute the value of P (Y > 3.75) as follows:

$P(Y>3.75)=\int\limits^{5}_{3.75} {\frac{1}{b-a}} \, dy \\\\=\int\limits^{5}_{3.75} {\frac{1}{5-1}} \, dy\\\\=\frac{1}{4}\times [y]^{5}_{3.75}\\\\=\frac{5-3.75}{4}\\\\=0.3125\\\\\approx 0.31$

Thus, the light will flash more than 3.75 seconds after the subject clicks "Start" 31% of the times.

(c)

Compute the 38th percentile as follows:

$P(Y$

Thus, the 38th percentile is 2.52 seconds.

4 0

3 years ago

What is the solution to the linear equation?

Vikki [24]

Answer:

x=-5

Step-by-step explanation:

7 0

4 years ago

Please help solve these proofs asap!!!

timama [110]

Answer:

Proofs contained within the explanation.

Step-by-step explanation:

These induction proofs will consist of a base case, assumption of the equation holding for a certain unknown natural number, and then proving it is true for the next natural number.

Proof

Base case:

We want to shown the given equation is true for n=1:

The first term on left is 2 so when n=1 the sum of the left is 2.

Now what do we get for the right when n=1:

$\frac{1}{2}(1)(3(1)+1)$

$\frac{1}{2}(3+1)$

$\frac{1}{2}(4)$

$2$

So the equation holds for n=1 since this leads to the true equation 2=2:

We are going to assume the following equation holds for some integer k greater than or equal to 1:

$2+5+8+\cdots+(3k-1)=\frac{1}{2}k(3k+1)$

Given this assumption we want to show the following:

$2+5+8+\cdots+(3(k+1)-1)=\frac{1}{2}(k+1)(3(k+1)+1)$

Let's start with the left hand side:

$2+5+8+\cdots+(3(k+1)-1)$

$2+5+8+\cdots+(3k-1)+(3(k+1)-1)$

The first k terms we know have a sum of .5k(3k+1) by our assumption.

$\frac{1}{2}k(3k+1)+(3(k+1)-1)$

Distribute for the second term:

$\frac{1}{2}k(3k+1)+(3k+3-1)$

Combine terms in second term:

$\frac{1}{2}k(3k+1)+(3k+2)$

Factor out a half from both terms:

$\frac{1}{2}[k(3k+1)+2(3k+2]$

Distribute for both first and second term in the [ ].

$\frac{1}{2}[3k^2+k+6k+4]$

Combine like terms in the [ ].

$\frac{1}{2}[3k^2+7k+4$

The thing inside the [ ] is called a quadratic expression. It has a coefficient of 3 so we need to find two numbers that multiply to be ac (3*4) and add up to be b (7).

Those numbers would be 3 and 4 since

3(4)=12 and 3+4=7.

So we are going to factor by grouping now after substituting 7k for 3k+4k:

$\frac{1}{2}[3k^2+3k+4k+4]$

$\frac{1}{2}[3k(k+1)+4(k+1)]$

$\frac{1}{2}[(k+1)(3k+4)]$

$\frac{1}{2}(k+1)(3k+4)$

$\frac{1}{2}(k+1)(3(k+1)+1)$ .

Therefore for all integers n equal or greater than 1 the following equation holds:

$2+5+8+\cdots+(3n-1)=\frac{1}{2}n(3n+1)$

Proof:

Base case: When n=1, the left hand side is 1.

The right hand at n=1 gives us:

$\frac{1}{4}(5^1-1)$

$\frac{1}{4}(5-1)$

$\frac{1}{4}(4)$

$1$

So both sides are 1 for n=1, therefore the equation holds for the base case, n=1.

We want to assume the following equation holds for some natural k:

$1+5+5^2+\cdots+5^{k-1}=\frac{1}{4}(5^k-1)$ .

We are going to use this assumption to show the following:

$1+5+5^2+\cdots+5^{(k+1)-1}=\frac{1}{4}(5^{k+1}-1)$

Let's start with the left side:

$1+5+5^2+\cdots+5^{(k+1)-1}$

$1+5+5^2+\cdots+5^{k-1}+5^{(k+1)-1}$

We know the sum of the first k terms is 1/4(5^k-1) given by our assumption:

$\frac{1}{4}(5^k-1)+5^{(k+1)-1}$

$\frac{1}{4}(5^k-1)+5^k$

Factor out the 1/4 from both of the two terms:

$\frac{1}{4}[(5^k-1)+4(5^k)]$

$\frac{1}{4}[5^k-1+4\cdot5^k]$

Combine the like terms inside the [ ]:

$\frac{1}{4}(5 \cdot 5^k-1)$

Apply law of exponents:

$\frac{1}{4}(5^{k+1}-1)$

Therefore the following equation holds for all natural n:

$1+5+5^2+\cdots+5^{n-1}=\frac{1}{4}(5^n-1)$ .

5 0

3 years ago

Alma is estimating the proportion of students in her school district who, in the past month, read at least 1 book. From a random

Maurinko [17]

Answer:

$(0.64 +1.645 \sqrt{\frac{0.64(0.36)}{50} })$

90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month

(0.52846 , 0.75166)

Step-by-step explanation:

Step:-1

Given that the random sample size 'n' = 50

The sample proportion

$p^{-} = \frac{32}{50} = 0.64$

90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month

$(p^{-} - Z_{0.90} \sqrt{\frac{p(1-p)}{n}, } , p^{-} +Z_{0.90} \sqrt{\frac{p(1-p)}{n} })$

Step(ii):-

Level of significance = 0.90

Z₀.₉₀ = 1.645

$(0.64 - 1.645 \sqrt{\frac{0.64(1-0.64)}{50}, } , 0.64 +1.645 \sqrt{\frac{0.64(1-0.64)}{50} })$

$(0.64 - 1.645 \sqrt{\frac{0.64(0.36)}{50}, } , 0.64 +1.645 \sqrt{\frac{0.64(0.36)}{50} })$

(0.64 -1.645(0.06788) , (0.64 + 1.645(0.06788)

(0.52846 , 0.75166)

Final answer:-

90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month

(0.52846 , 0.75166)

5 0

3 years ago