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3 years ago

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Sara can take no more than 22 pounds of luggage on a trip. Her suitcase weighs 112 ounces, How many more pounds can she pack wit

hout going over the limit?
Show me your work please. :) <3

1 answer:

postnew [5]3 years ago

3 0

OK so 1 pound=16 ounces
so 112 divided by 16 is 7
22-7= 15
may be wrong with calculation I did mental math but there is the work just check answers

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What does mx+b stand for

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Answer:

thats the equation for slope-intercept form

Step-by-step explanation:

m is the slope and b is where the line intercepts on the y axis

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3 years ago

Read 2 more answers

Scores of an IQ test have a bell-shaped distribution with a mean of 100 and a standard deviation of 17. Use the empirical rule

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Answer:

(a) 95% of people has an IQ score between 66 and 134.

(b) 5% of people has an IQ score less than 66 or greater than 134.

(c) 2.5% of people has an IQ score greater than 134.

Step-by-step explanation:

We are given that Scores of an IQ test have a bell-shaped distribution with a mean of 100 and a standard deviation of 17.

Let X = <u><em>Scores of an IQ test </em></u>

So, X ~ Normal( $\mu=100, \sigma^{2} = 17^{2}$ )

Now, the Empirical rule states that;

68% of the data values lies within one standard deviation of the means.
95% of the data values lies within two standard deviation of the means.
99.7% of the data values lies within three standard deviation of the means.

That is; [ $\mu-\sigma,\mu+\sigma$ ] = [ $100-17,100+17$ ]

= $[83,117]$

[ $\mu-2\sigma,\mu+2\sigma$ ] = [ $100-34,100+34$ ]

= $[66,134]$

[ $\mu-3\sigma,\mu+3\sigma$ ] = [ $100-51,100+51$ ]

= $[49,151]$

(a) Percentage of people that has an IQ score between 66 and 134 is given by = P(66 < X < 134)

As seen above this value lies in the second category which means that 95% of people that has an IQ score between 66 and 134.

(b) Percentage of people that has an IQ score less than 66 or greater than 134 is given by;

As we know that 95% of the data values lies within 66 and 134, so the percentage of people that has an IQ score less than 66 or greater than 134 is = 100% - 95% = 5%.

(c) Since it has been calculated above that 5% of people has an IQ score less than 66 or greater than 134 which means half of these people will lie below score of 66 and half of these will lie above score of 134.

SO, percentage of people that has an IQ score greater than 134 = $\frac{5\%}{2}$ = 2.5%

Hence, 2.5% of people has an IQ score greater than 134.

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3 years ago

Assume {v1, . . . , vn} is a basis of a vector space V , and T : V ------> W is an isomorphism where W is another vector spac

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Answer:

Step-by-step explanation:

To prove that $w_1,\dots w_n$ form a basis for W, we must check that this set is a set of linearly independent vector and it generates the whole space W. We are given that T is an isomorphism. That is, T is injective and surjective. A linear transformation is injective if and only if it maps the zero of the domain vector space to the codomain's zero and that is the only vector that is mapped to 0. Also, a linear transformation is surjective if for every vector w in W there exists v in V such that T(v) =w

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$\lambda_1w_1+\dots \lambda_n w_n=0$ implies that

$\lambda_1 = \cdots = \lambda_n$ .

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$\lambda_1w_1+\dots \lambda_n w_n=0$

If we apply $T^{-1}$ to this equation, then, we get

$T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) =T^{-1}(0) = 0$

Since T is linear, its inverse is also linear, hence

$T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) = \lambda_1T^{-1}(w_1)+\dots + \lambda_nT^{-1}(w_n)=0$

which is equivalent to the equation

$\lambda_1v_1+\dots + \lambda_nv_n =0$

Since $v_1,\dots,v_n$ are linearly independt, this implies that $\lambda_1=\dots \lambda_n =0$ , so the set $\{w_1, \dots, w_n\}$ is linearly independent.

Now, we will prove that this set generates W. To do so, let w be a vector in W. We must prove that there exist $a_1, \dots a_n$ such that

$w = a_1w_1+\dots+a_nw_n$

Since T is surjective, there exists a vector v in V such that T(v) = w. Since $v_1,\dots, v_n$ is a basis of v, there exist $a_1,\dots a_n$ , such that

$a_1v_1+\dots a_nv_n=v$

Then, applying T on both sides, we have that

$T(a_1v_1+\dots a_nv_n)=a_1T(v_1)+\dots a_n T(v_n) = a_1w_1+\dots a_n w_n= T(v) =w$

which proves that $w_1,\dots w_n$ generate the whole space W. Hence, the set $\{w_1, \dots, w_n\}$ is a basis of W.

Consider the linear transformation $T:\mathbb{R}^2\to \mathbb{R}^2$ , given by T(x,y) = T(x,0). This transformations fails to be injective, since T(1,2) = T(1,3) = (1,0). Consider the base of $\mathbb{R}^2$ given by (1,0), (0,1). We have that T(1,0) = (1,0), T(0,1) = (0,0). This set is not linearly independent, and hence cannot be a base of $\mathbb{R}^2$

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3 years ago

In which direction does the graph of the function shown below open?

sp2606 [1]

Answer:

D. Up

Step-by-step explanation:

When a parabola has the form $y=ax^2+bx+c$ , It is vertical (opens up or down).

Because the variable "x" is squared.

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$a=2$

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