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mestny [16]
3 years ago
5

I need help plz plz plz

Mathematics
2 answers:
posledela3 years ago
6 0

solving eqn 1st

<u>x</u><u> </u>=<u>2</u>

y. 3

3x=2y

x =<u>2</u><u>y</u>

<u> </u><u> </u><u> </u><u> </u><u> </u>3

again putting the value of y in eqn 1st

x= <u>2</u><u>y</u>

<u> </u><u> </u><u> </u><u> </u><u> </u>3

x= <u>2</u><u>×</u><u>9</u>

<u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>3

x=<u>1</u><u>8</u>

<u> </u><u> </u><u> </u><u> </u>3

x =6

now,

solving, eqn 2nd

<u>y</u><u> </u>= <u>3</u>

12. 4

4y =36

y=<u>3</u><u>6</u>

<u> </u><u> </u><u> </u><u> </u><u> </u>4

y=9

hence,

the value of x =6 and y= 9.

hope it help u.

AfilCa [17]3 years ago
3 0

Answer:

12/18 = 2/3

9/12 = 3/4

sorry if this doesn’t help!

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dybincka [34]

Answer:

The given lines are perpendicular.

Step-by-step explanation:

In the question, two lines are given as line 1 passes through (1,7) and (5,5). Whereas, line 2 passes through (-1,-3) and (1,1).

It is required to find the slope of given lines and figure out whether they are perpendicular, parallel or neither.

To solve this question, first find the slope of both lines. Check if their product is equal to -1, then they are perpendicular. If the slopes are equal the lines are parallel.

Step 1 of 2

Find the slope of first line.

$$\begin{aligned}m_{1} &=\frac{5-7}{5-1} \\m_{1} &=\frac{-2}{4} \\m_{1} &=-\frac{1}{2}\end{aligned}$$

Step 2 of 2

Find the slope of first line.

$$\begin{aligned}&m_{2}=\frac{1-(-3)}{1-(-1)} \\&m_{2}=\frac{4}{2} \\&m_{2}=2\end{aligned}$$

And

$$\begin{aligned}&m_{1} m_{2}=-\frac{1}{2}(2) \\&m_{1} m_{2}=-1\end{aligned}$$

Since, both slopes are reciprocal of each other.

Therefore, the lines are perpendicular.

6 0
2 years ago
A company employs two shifts of workers. Each shift produces a type of gasket where the thickness is the critical dimension. The
Oksanka [162]

Answer:

a) [-0.134,0.034]

b) We are uncertain

c) It will change significantly

Step-by-step explanation:

a) Since the variances are unknown, we use the t-test with 95% confidence interval, that is the significance level = 1-0.05 = 0.025.

Since we assume that the variances are equal, we use the pooled variance given as

s_p^2 = \frac{ (n_1 -1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2},

where n_1 = 40, n_2 = 30, s_1 = 0.16, s_2 = 0.19.

The mean difference \mu_1 - \mu_2 = 10.85 - 10.90 = -0.05.

The confidence interval is

(\mu_1 - \mu_2) \pm t_{n_1+n_2-2,\alpha/2} \sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_2}} = (-0.05) \pm t_{68,0.025} \sqrt{\frac{0.03}{40} + \frac{0.03}{30}}

= -0.05\pm 1.995 \times 0.042 = -0.05 \pm 0.084 = [-0.134,0.034]

b) With 95% confidence, we can say that it is possible that the gaskets from shift 2 are, on average, wider than the gaskets from shift 1, because the mean difference extends to the negative interval or that the gaskets from shift 1 are wider, because the confidence interval extends to the positive interval.

c) Increasing the sample sizes results in a smaller margin of error, which gives us a narrower confidence interval, thus giving us a good idea of what the true mean difference is.

6 0
4 years ago
= 8 cm, height = 67 cm.
zavuch27 [327]

Answer:

45

Step-by-step explanation:

8cm=67-8

5 0
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vivado [14]

Answer:

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i took the test hope this helped

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3 years ago
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Afina-wow [57]

1.25

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