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3 years ago

1 4 3 In the image above, 23 = 129, what is the value of 22?

Mathematics

1 answer:

mylen [45]3 years ago

7 0

Answer:

where's the image given bro

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A blueprint for a house shows the height of the house to be 3 3/8 inches. The actual house is 21 1/3 feet tall. What is the unit

Amanda [17]

6.32ft per inch (rounded to the hundredth place).

You can get this number by dividing the total number of feet by the total number of inches

6 0

4 years ago

On a coordinate plane, triangle A B C is shifted 4 units to the right and 2 units up to form triangle A prime B prime C prime. I

zheka24 [161]

Answer:

(x, y) ⟶ (x + 4, y + 2)

Step-by-step explanation:

If you shift a point four units to the right, x ⟶ x + 4.

If you shift it up two units, y ⟶ y + 2.

If you combine the two shifts into one transformation, the rule becomes

(x, y) ⟶ (x + 4, y + 2)

3 0

3 years ago

Ron's piano lesson starts in 2 hours. It takes him 30 minutes to get to the lesson. He has

RideAnS [48]

Answer:

2 hours=120 mins

120-(45+25)

120-(70)

Ron has enough time to finish his homework and walk the dog before his lesson. Ron should overestimate the total time needed. Since Ron has 50 minutes to spare he could add 25 to home work time and another 25 to walking his dog. he still ha time to do both.

4 0

3 years ago

Activity 4: Performance Task

Nookie1986 [14]

An arithmetic progression is simply a progression with a common difference among consecutive terms.

The sum of multiplies of 6 between 8 and 70 is 390
The sum of multiplies of 5 between 12 and 92 is 840
The sum of multiplies of 3 between 1 and 50 is 408
The sum of multiplies of 11 between 10 and 122 is 726
The sum of multiplies of 9 between 25 and 100 is 567
The sum of the first 20 terms is 630
The sum of the first 15 terms is 480
The sum of the first 32 terms is 3136
The sum of the first 27 terms is -486
The sum of the first 51 terms is 2193

(a) Sum of multiples of 6, between 8 and 70

There are 10 multiples of 6 between 8 and 70, and the first of them is 12.

This means that:

$\mathbf{a = 12}$

$\mathbf{n = 10}$

$\mathbf{d = 6}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{10} = \frac{10}2(2*12 + (10 - 1)6)}$

$\mathbf{S_{10} = 390}$

(b) Multiples of 5 between 12 and 92

There are 16 multiples of 5 between 12 and 92, and the first of them is 15.

This means that:

$\mathbf{a = 15}$

$\mathbf{n = 16}$

$\mathbf{d = 5}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{16} = \frac{16}2(2*15 + (16 - 1)5)}$

$\mathbf{S_{16} = 840}$

(c) Multiples of 3 between 1 and 50

There are 16 multiples of 3 between 1 and 50, and the first of them is 3.

This means that:

$\mathbf{a = 3}$

$\mathbf{n = 16}$

$\mathbf{d = 3}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{16} = \frac{16}2(2*3 + (16 - 1)3)}$

$\mathbf{S_{16} = 408}$

(d) Multiples of 11 between 10 and 122

There are 11 multiples of 11 between 10 and 122, and the first of them is 11.

This means that:

$\mathbf{a = 11}$

$\mathbf{n = 11}$

$\mathbf{d = 11}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{16} = \frac{11}2(2*11 + (11 - 1)11)}$

$\mathbf{S_{11} = 726}$

(e) Multiples of 9 between 25 and 100

There are 9 multiples of 9 between 25 and 100, and the first of them is 27.

This means that:

$\mathbf{a = 27}$

$\mathbf{n = 9}$

$\mathbf{d = 9}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{9} = \frac{9}2(2*27 + (9 - 1)9)}$

$\mathbf{S_{9} = 567}$

(f) Sum of first 20 terms

The given parameters are:

$\mathbf{a = 3}$

$\mathbf{d = 3}$

$\mathbf{n = 20}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{20} = \frac{20}2(2*3 + (20 - 1)3)}$

$\mathbf{S_{20} = 630}$

(f) Sum of first 15 terms

The given parameters are:

$\mathbf{a = 4}$

$\mathbf{d = 4}$

$\mathbf{n = 15}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{15} = \frac{15}2(2*4 + (15 - 1)4)}$

$\mathbf{S_{15} = 480}$

(g) Sum of first 32 terms

The given parameters are:

$\mathbf{a = 5}$

$\mathbf{d = 6}$

$\mathbf{n = 32}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{32} = \frac{32}2(2*5 + (32 - 1)6)}$

$\mathbf{S_{32} = 3136}$

(g) Sum of first 27 terms

The given parameters are:

$\mathbf{a = 8}$

$\mathbf{d = -2}$

$\mathbf{n = 27}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{27} = \frac{27}2(2*8 + (27 - 1)*-2)}$

$\mathbf{S_{27} = -486}$

(h) Sum of first 51 terms

The given parameters are:

$\mathbf{a = -7}$

$\mathbf{d = 2}$

$\mathbf{n = 51}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{51} = \frac{51}2(2*-7 + (51 - 1)*2)}$

$\mathbf{S_{51} = 2193}$

Read more about arithmetic progressions at:

brainly.com/question/13989292

4 0

2 years ago

Read 2 more answers

If t follows a t7 distribution, find t0 such that (a) p(|t | < t0) = .9 and (b) p(t > t0) = .05.

Thepotemich [5.8K]

Answer:

a) $P(-t_o < t_7$

Using the symmetrical property we can write this like this:

$1-2P(t_7$

We can solve for the probability like this:

$2P(t_7$

$P(t_7$

And we can find the value using the following excel code: "=T.INV(0.05,7)"

So on this case the answer would be $t_o =\pm 1.895$

b) For this case we can use the complement rule and we got:

$1-P(t_7$

We can solve for the probability and we got:

$P(t_7$

And we can use the following excel code to find the value"=T.INV(0.95;7)"

And the answer would be $t_o = 1.895$

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

For this case we know that $t \sim t(df=7)$

And we want to find:

Part a

$P(|t|$

We can rewrite the expression using properties for the absolute value like this:

$P(-t_o < t_7$

Using the symmetrical property we can write this like this:

$1-2P(t_7$

We can solve for the probability like this:

$2P(t_7$

$P(t_7$

And we can find the value using the following excel code: "=T.INV(0.05,7)"

So on this case the answer would be $t_o =\pm 1.895$

Part b

$P(t>t_o) =0.05$

For this case we can use the complement rule and we got:

$1-P(t_7$

We can solve for the probability and we got:

$P(t_7$

And we can use the following excel code to find the value"=T.INV(0.95;7)"

And the answer would be $t_o = 1.895$

5 0

3 years ago