MATHEMATICS COLLEGE

Marie plants 12 packages of vegetable seeds in a community garden. Each package costs $1.97. What is the total cost of the seeds?

Answers

Answer 1

Answer:

Answer:

$23.64

Step-by-step explanation:

12 * $1.97 = $23.64

Related Questions

jewelry box with a square base is to be built with silver plated sides, nickel plated bottom and top, and a volume of 44 cm3. If nickel plating costs $1 per cm2 and silver plating costs $2 per cm2, find the dimensions of the box to minimize the cost of the materials. (Round your answers to two decimal places.) The box which minimizes the cost of materials has a square base of side length cm and a height of cm.
What is the reciprocal of 5/6
Can someone please answer asap..ik it’s easy but idk lol
The recommended dietary allowance (RDA) of iron for adult females is 18 milligrams (mg) per day. The given iron intakes (mg) were obtained for 45 random adult females. At the 10% significance level, do the data suggest that adult females are, on average, getting less than the RDA of 18 mg of iron? Assume that the population standard deviation is 4.2 mg. Preliminary data analyses indicate that applying the z-test is reasonable. (Note: x bar equals=14.65 mg.)1. State the hypotheses for the one-mean z-test.2. Compute the value of the test statistic. z=______3. Determine the critical value(s).4._________ the null hypothesis. At the 10% significance level, the data____________sufficient evidence to conclude that adult females are ___________ the RDA of iron, on average.
Order the following numbers from least to greatest

Work out the height of this cone

Answers

Answer:20 cm

Step-by-step explanation:

Volume of cone=540π

Radius=r=9

Volume of cone=1/3 x π x r^2 x h

540π=1/3 x π x 9^2 x h

540π=1/3 x π x 9 x 9 x h

540π=(1xπx9x9xh)/3

540π=(81πh)/3

540π=27πh

Divide both sides by 27π

540π/27π=(27πh)/27π

20=h

h=20

Height =20 cm

A giraffe is 5 m 20cm tall. An Elephant is 1m 77cm shorter than the giraffe. A rhinoceros is 1m 58 cm shorter than the elephant. How tall is the rhinoceros

Answers

The answer 1m and 85cm

A coin is tossed and a fourth section spinner is spun once a tree diagram shows the possible outcomes for the two events what is the probability of flipping tails on a coin and spinning a two on the spinner

Answers

Answer:

2/6 or 1/3 I think

Step-by-step explanation:

you have a 1 out of 2 chances to land tails and 1 out of 4 chance to role a 2 on the spinner. so in total there is 6 chances for rolling and flipping the coin. and there is only 1 chance for both in the odds.

therefore I believe that its is 1/3 or 2/6. I do not know if I am right exactly but that is my thought process.

Will Mark Brainlest help please

Answers

Answer:

g(-1 )=-1 and g(2)+g(1)=7

Step-by-step explanation:

If g(x) = x^3+x^2-x-2 find g(-1)

if we find g(-1)

we substitute all the x's in the function with -1

-1^3+-1^2-(-1)-2

-1^3 = -1

-1^2 = 1

-1+1+1-2

(two minuses make a plus)

-1+1 = 0

0+1 = 1

1-2 = -1

if x=-1, g(-1) is -1

g(2)+g(1)

substitute the x's in the function with 2 and 1 and add your results

2^3+2^2-2-2

2^3 = 8, 2^2 = 4

8+4-2-2

8+4= 12, 12-2 = 10, 10-2 = 8

g(2)=8

g(1) now

1^3 + 1^2-1-2

1^3=1, 1^2 = 1

1+1-1-2

1+1 = 2, 2-1 = 1, 1-2 = -1

g(3) = -1

g(2) (which equals 8) + g(3) (which equals -1) =

8+(-1) = 7

g(2)+g(3)=7

Use the Gram-Schmidt process to find an orthonormal basis for the subspace of R4 spanned by the vectors (1, 0, 1, 1), (1, 0, 1, 0), (0, 0, 1, 1).

Answers

Answer:

$$ e_1 = \begin{pmatrix}\frac{\sqrt{\textbf{3}}}{\textbf{2}}\n\n\textbf{0} \n\n\frac{\sqrt{\textbf{3}}}{\textbf{3}}\n\n\frac{\sqrt{\textbf{3}}}{\textbf{3}}\end{pmatrix}$ $$ e_2 = \begin{pmatrix}\frac{\sqrt{\textbf{6}}}{\textbf{6}}\n\n\textbf{0} \n\n\frac{\sqrt{\textbf{6}}}{\textbf{6}}\n\n\frac{\sqrt{\textbf{-6}}}{\textbf{3}}\end{pmatrix}$ $$ e_3 = \begin{pmatrix}\frac{\sqrt{\textbf{-2}}}{\textbf{2}}\n\n\textbf{0} \n\n\frac{\sqrt{\textbf{2}}}{\textbf{2}}\n\n0\end{pmatrix}$

Step-by-step explanation:

we have to orthonormalize the vectors:

$v_1 = \begin{pmatrix} 1 \n 0 \n 1 \n 1 \end{pmatrix}$ $v_2 = \begin{pmatrix} 1 \n 0 \n 1 \n 0 \end{pmatrix}$ $$ v_3 = \begin{pmatrix} 0 \n 0 \n 1 \n 1 \end{pmatrix}$

According to Gram - Schmidt process, we have:

$u_k = v_k - \sum_(j = 1) ^ {k - 1} proj_(uj) (v_k)$ where, $$ proj_u (v) = (u . v)/(u . u)u$

The normalized vector is: $e_k = (u_k)/(√(u_k.u_k))$

Now, the first step.

$v_1 = \begin{pmatrix} 1 \n 0 \n 1 \n 1 \end{pmatrix}$ = u₁

Therefore, e₁ = $(u_1)/(√(u_1.u_1))$

$$ = \begin{pmatrix}\frac{\sqrt{\textbf{3}}}{\textbf{2}}\n\n\textbf{0} \n\n\frac{\sqrt{\textbf{3}}}{\textbf{3}}\n\n\frac{\sqrt{\textbf{3}}}{\textbf{3}}\end{pmatrix}$

Now, we find e₂.

$u_2 = v_2 - (u_1.v_2)/(u_1.u_1)u_1$

$$ = \begin{pmatrix} (1)/(3)\n \n 0 \n\n (1)/(3)\n\n (-2)/(3) \end{pmatrix}$

Therefore, $e_2 = (u_2)/(√(u_2.u_2))$

$$ e_2 = \begin{pmatrix}\frac{\sqrt{\textbf{6}}}{\textbf{6}}\n\n\textbf{0} \n\n\frac{\sqrt{\textbf{6}}}{\textbf{6}}\n\n\frac{\sqrt{\textbf{-6}}}{\textbf{3}}\end{pmatrix}$

To find e₃:

$u_3 = v_3 - (u_1. v_3)/(u_1.u_1)u_1 - (u_2. v_3)/(u_2.u_2) u_2$

$$ = \begin{pmatrix} (-1)/(2) \n\n 0\n \n (1)/(2) \n\n 0 \n\end{pmatrix}$

$e_3 = (u_3)/(√(u_3.u_3))$

$$ e_3 = \begin{pmatrix}\frac{\sqrt{\textbf{-2}}}{\textbf{2}}\n\n\textbf{0} \n\n\frac{\sqrt{\textbf{2}}}{\textbf{2}}\n\n0\end{pmatrix}$

So, we have the orthonormalized vectors $e_1, e_2, e_3$ .

Hence, the answer.

Final answer:

To find the orthonormal basis using the Gram-Schmidt process, we calculate the first vector by dividing the first given vector by its magnitude and normalize it. Then, we subtract the projection of each subsequent vector onto the previously found orthonormal vectors and normalize the resulting vector.

Explanation:

To find an orthonormal basis for the subspace of R4 spanned by the given vectors using the Gram-Schmidt process, we will start by finding the first vector of the orthonormal basis. Let's call the given vectors v1, v2, and v3, respectively. The first vector of the orthonormal basis, u1, is equal to v1 divided by its magnitude, which is ||v1||. So, u1 = v1 / ||v1||. We can calculate ||v1|| as √(1^2 + 0^2 + 1^2 + 1^2) = √3.

Therefore, u1 = (1/√3, 0/√3, 1/√3, 1/√3).

Now, we need to find u2, the second vector of the orthonormal basis. To do this, we subtract the projection of v2 onto u1 from v2, then divide the result by its magnitude. We calculate the projection of v2 onto u1 as proj_u1(v2) = u1 * dot(u1, v2), where dot(u1, v2) represents the dot product of u1 and v2.

Finally, we subtract proj_u1(v2) from v2 to get v2' = v2 - proj_u1(v2), and then normalize v2' to get u2 = v2' / ||v2'||.

We can repeat this process to find u3, the third vector of the orthonormal basis. Subtract proj_u1(v3) and proj_u2(v3) from v3, then normalize the result to get u3 = v3' / ||v3'||.

Therefore, the orthonormal basis for the subspace spanned by the given vectors is (1/√3, 0/√3, 1/√3, 1/√3), (0, 0, 0, 1), and (-1/√3, 0/√3, 1/√3, 1/√3).

Learn more about Gram-Schmidt process here:

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Determine the x-intercept of the line whose equation is given:y = StartFraction x Over 2 EndFraction minus 3
a.
(6, 0)
b.
(negative 6, 0)
c.
(0, three-halves)
d.
(Negative three-halves, 0)