Complete the statement. [Type your answer as a number.]<br><br> 7 pounds × 16 ounces

Hi anyone can help me do this question?

lakkis [162]

Answer:

6+18=24+16=40 Top one 3+3=6+8=14

Step-by-step explanation:

Easy by adding

5 0

3 years ago

(1) Ax + By = C (2) Dx + Ey = F Jim is trying to solve the system of equations. He begins by multiplying equation (1) by D and e

attashe74 [19]

Answer:

No, Angela is not correct.

As the method Jim was performing will also lead to the solution.

Step-by-step explanation:

We are given first equation as:

Ax + By = C

Second equation is:

Dx + Ey = F

Jim solved the equation as:

He begins by multiplying equation (1) by D and equation (2) by A.

and so by subtracting both the equations he will obtain a value of y.

and then put the y-value in any of the given two equations to obtain the value of x.

Angela Method:

you should have multiplied equation (1) by E and equation (2) by B.

and when she will subtract both the equations she will get the value of x first and then when she will put the value of x in any of the given equation she will obtain the value of y.

But both will get a value for x and y.

Hence, the method Jim was performing was also correct.

8 0

3 years ago

Read 2 more answers

The school system is trying to help their students be more successful. They are going to hire tutors to help. They must have mor

dexar [7]

Answer:

a. $x-y > 0$

b. $x+y \leq 50$

c. x=26; y=24 or (24,26)

Step-by-step explanation:

Let x = number of teachers hired.

Let y = number of tutors hired.

Now solving for part a we get

a. Write an inequality that represents the statement that the number of teachers hired must exceed the number of tutors hired.

$x-y > 0$

solving for part b we get;

b. Write an inequality that represents the statement that the maximum number of teachers and tutors is 50.

$x+y \leq 50$

solving for part c we get;

c. Choose a point that satisfies the situation, and explain why you chose that number of tutors and teachers.

Now we know that $x-y > 0$ also $x+y \leq 50$

x=26; y=24

(24,26)

Explanation: To make number of teacher more than number of tutors this is the maximum value we can achieve for the requirement.

6 0

4 years ago

How many blocks with dimensions of One-third times 1 times 1 can fit in a unit cube?

lord [1]

Given:

Dimensions of a block are $\dfrac{1}{3}\times 1\times 1$ .

To find:

The number of block that can be fit in a unit cube.

Solution:

Volume of a cuboid is:

$V=l\times b\times h$

Where, l is length, b is breadth or width and h is the height of the cuboid.

So, the volume of the given block is:

$V_1=\dfrac{1}{3}\times 1\times 1$

$V_1=\dfrac{1}{3}$

Dimensions of a unit cube are $1\times 1\times 1$ . So, the volume of the unit cube is:

$V_2=1\times 1\times 1$

$V_2=1$

We need to divide the volume of unit cube by the volume of a block to find the number of block that can be fit in a unit cube.

So, the number of blocks that fit in a unit cube is:

$n=\dfrac{V_2}{V_1}$

$n=\dfrac{1}{\dfrac{1}{3}}$

$n=3$

Therefore, the correct option is B.

8 0

3 years ago

Read 2 more answers

In order to justify this claim for the Ferengi a sample of 16 coils is taken. For this sample set, the mean warp phase flux is f

Kay [80]

Answer:

The responses to these question can be defined as follows:

Step-by-step explanation:

$n = 16\\\\ \bar{x}= 73.7\\\\\sigma = 12\\\\a = 0.05\ or\ a = 0.10\\\\H_{o} \ : \mu = 68\\\\H_{a} \ : \mu \neq 68\\\\a = 0.05\\\\$

critical values $=\pm t0.025,15 = \pm 2.131\\\\$

$(n-1) = 15^{\circ}\\\\a = 0.10\\\\$

critical values $= \pm t0.05,15 = \pm 1.753\\\\$

$(n-1) = 15^{\circ}\\\\$

Testing the statistic values:

$t = \frac{x-\mu_{0}}{ \frac{s}{\sqrt{n}}}\\\\$

$= \frac{73.7-68}{(\frac{12}{\sqrt{16}})}\\\\\ = \frac{5.7}{(\frac{12}{4})}\\\\ = \frac{5.7}{(3)}\\\\ = 1.9\\$

Test statistic ta $= -1.90\ lies$

The critical values $\pm t_{0.05,15} =\pm 1.753$

It is in the region of dismissal. We dismiss the 10% significant null hypothesis.

$t_a = 1.90 \\\\df = 15\\\\a = 0.05\\\\p-value = 076831\\\\$

P - value is greater than the level of significance a= 0.05

Null hypothesis we don't reject. At a 95% level, the claim is justified.

$t_a = 1.90\\\\ df = 15\\\\ a = 0.10\\\\p-value = 076831\\\\$

P - value below the meaning level a = 0.10, we reject the hypothesis null. At a level of 90% the claim is not justified.

3 0

3 years ago

Complete the statement. [Type your answer as a number.] 7 pounds × 16 ounces