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2 years ago

Pls help me answer this it's due today

Mathematics

2 answers:

kvv77 [185]2 years ago

6 0

Answer:

I think its 21 But if its times then its 240

Step-by-step explanation:

Hope this helped

Sonbull [250]2 years ago

4 0

Answer:

120

Step-by-step explanation:

formula for volume is area×height

v=1 /2b×h×l

v=1/2×5×4×12

v=2.5×4×12

v=10×12

$v= {120cm}^{3}$

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How do you know that the decimal 2\11 repeats without end?

seropon [69]

It doesn't, it ends after 32 digits after 0.

5 0

3 years ago

Determine

mihalych1998 [28]

Answer:

$=4.+2.$

Step-by-step explanation:

<u>Linear Combination Of Vectors </u>

One vector $\vec b$ is a linear combination of $\vec a_1$ and $\vec a_2$ if there are two scalars $x_1, x_2$ such as

$\vec b=x_1\vec a_1+x_2\vec a_2$

In our case, all the vectors are given in $R^3$ but there are only two possible components for the linear combination. This indicates that only two conditions can be used to determine both scalars, and the other condition must be satisfied once the scalars are found.

We have

$\vec a_1=,\ \vec a2=,\ \vec b=$

We set the equation

$=x_1.+x_2.$

Multiplying both scalars by the vectors

$=+$

Equating each coordinate, we get

$4x_1-4x_2=8$

$5x_1+3x_2=26$

$-4x_1+3x_2=-10$

Adding the first and the third equations:

$-x_2=-2$

$x_2=2$

Replacing in the first equation

$4x_1-4(2)=8$

$4x_1=8+8$

$x_1=4$

We must test if those values make the second equation become an identity

$5(4)+3(2)=20+6=26$

The second equation complies with the values of $x_1$ and $x_2$ , so the solution is

$=4.+2.$

8 0

3 years ago

A company claims that the mean weight per apple they ship is 120 grams with a standard deviation of 12 grams. Data generated fro

Veronika [31]

Answer:

There is no sufficient evidence to reject the company's claim at the significance level of 0.05

Step-by-step explanation:

Let $\mu$ be the true mean weight per apple the company ship. We want to test the next hypothesis

$H_{0}:\mu=120$ vs $H_{1}:\mu\neq 120$ (two-tailed test).

Because we have a large sample of size n = 49 apples randomly selected from a shipment, the test statistic is given by

$Z=\frac{\bar{X}-120}{\sigma/\sqrt{n}}$ which is normally distributed. The observed value is

$z_{0}=\frac{122.5-120}{12/\sqrt{49}}=1.4583$ . The rejection region for $\alpha = 0.05$ is given by RR = {z| z < -1.96 or z > 1.96} where the area below -1.96 and under the standard normal density is 0.025; and the area above 1.96 and under the standard normal density is 0.025 as well. Because the observed value 1.4583 does not fall inside the rejection region RR, we fail to reject the null hypothesis.