Between June 2017 and June 2018, the price of gasoline increased by 25%. If the price

A 3-ounce serving of tuna provides 21 g of protein use equivalent ratios to find how many grams of protein are in 9 ounces of tu

cluponka [151]

Since 3 ounces=21 g of protein, you would multiply 21 g x 3, because 6 ounces would = 42g. using that principle 3×3 = 9 & 21×3 = 63.

4 0

3 years ago

Hi I just really need help on this

Komok [63]

Answer:

3. 9 free throws. Proportion: 9/15

4. 18 free throws. Proportion: 18/24

Step-by-step explanation:

Multiply the number of free throws by the decimal of the percentage:

15*60%=15*.60=9

18*75%=18*.75=18.

The proportion is the number of free throws made over the total number of free throws attempted.

Hope this helps! :)

Be sure to give me the brainliest answer, and add me on snap,@ metro013103, I do homework and give answers on there. :P

4 0

3 years ago

Of 560 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compared

frutty [35]

Answer:

a) (0.5256,0.5944)

c) Criticism is invalid

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 560

Proportion of mislabeled = 56%

$\hat{p} = 0.56$

a) 90% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.64$

Putting the values, we get:

$0.56\pm 1.64(\sqrt{\dfrac{0.56(1-0.56)}{560}}) = 0.56\pm 0.0344\\\\=(0.5256,0.5944)$

b) Interpretation of confidence interval:

We are 90% confident that the true proportion of all seafood in the country that is mislabeled or misidentified is between 0.5256 and 0.5944 that is 52.56% and 59.44%.

c) Validity of criticism

Conditions for validity:

$np > 10\\n(1-p)>10$

Verification:

$560\times 0.56 = 313.6>10\\560(1-0.56) = 246.4>10$

Both the conditions are satisfied. This, the criticism is invalid.

8 0

3 years ago

One month Reuben rented movies and video games for a total of . The next month he rented movies and video games for a total of .

Vesnalui [34]

Can you say the total cost? There's not enough to go by to answer this.

4 0

3 years ago

Which of the following subsets of ℝ3×3 are subspaces of ℝ3×3? A. The 3×3 matrices whose entries are all greater than or equal to

Debora [2.8K]

Answer:

A. It is NOT a subspace of R^3x3

B. It IS a subspace of R^3x3

C. It IS a subspace of R^3x3

D. It is NOT a subspace of R^3x3

E It is NOT a subspace of R^3x3

F. It IS a subspace of R^3x3

Step-by-step explanation:

A way to show that a set is not a subspace, it´s enough to show that some properties of the definition of a vector spaces does not apply in that set or that operations under that set are not closed (we can get out of the set with linear combinations of elements in the set).

A. For definition of subspace, we know that every element has to have an additive inverse, but in set "A" (The 3×3 matrices whose entries are all greater than or equal to 0 ) every entry is greater than or equal to zero. In this set, there´s no additive inverse with the usual sum in R^3x3.

If sufficient to prove a set is a subspace showing that zero is in the set, there are additive inverses and that operations (sum and scalar multiplication) are closed in that set.

B. Notice that the matrix 0 is in "B" (The 3×3 matrices A such that the vector (276) is in the kernel of A), also notice if A(276)=0 then -A(276)=0 so every additive inverse (of an element in "B") belongs to "B".

Now we just have to prove that operations are closed in "B". Let X,Y matrices in set "B" and let z a scalar from the field. We are going to show that:

zX+Y ∈ B

For definition of set B:

X(276)=0 and Y(276)=0

So for zX+Y:

(zX+Y)(276)=zX(276)+Y(276)=z(0)+(0)

(zX+Y)(276)=0

So (276) is in the kernel of zX+Y, i.e (zX+Y) ∈ B.

We conclude "B" (with usual sum and scalar product of matrices) is a subspace of R^3x3

C. Notice the matrix 0 ∈ "C" (The diagonal 3×3 matrices) and there are all the additive inverse of the elements in "C". With the usual sum and scalar product, if the only zero entries are above and under the diagonal, it´ll stay like that no matter what linear combination we do because sum of matrices is entry by entry, and for every entry above or under the diagonal the sum and scalar product of two elements is going to be 0 in the same entries under and above the diagonal. "C" is a subspace

D. In set "D" (The non-invertible 3×3 matrices) it´s necessary to show that the sum is not closed:

Consider the following matrices and their sum:

$X=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&0\end{array}\right]\\ Y=\left[\begin{array}{ccc}0&0&0\\0&0&0\\0&0&1\end{array}\right]$

X+Y=I

We showed that sum is not closed in "C", so "C" is not a subspace of R^3x3

E. The definition of a reduced row-echelon matrix requires that the first element of a row must be 1, but with sum and scalar multiplication is easy to show that these pivot could easily change its value. So the set "E" is not closed under the usual operations under R^3x3.

F. The argument is similar to part C. No matter what linear combination we do, the last row is always going to be zero (with the usual operations in R^3x3). 0 ∈ "F" (The 3×3 matrices with all zeros in the third row) and all additive inverses (for an element in "F") is in "F", we affirm that "F" is a subspace of R^3x3

5 0

2 years ago