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katen-ka-za [31]
3 years ago
10

What is the same in all of these polygons

Mathematics
2 answers:
sdas [7]3 years ago
6 0
My guess is that they all have four (4) sides because polygons only have 4 sides
MrMuchimi3 years ago
3 0
Please add a photo so we can understand what you mean and your question.
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I NEED HELP PLEASE! THANKS :)
Pavlova-9 [17]

Answer:

6000kg/m^3

10°c

you need a unit for 4.3

5 0
3 years ago
Which of the following is equivalent to the complex number i^9
alexgriva [62]

Answer:

i

Step-by-step explanation:

i9=

(i4)2×i1=

(1)2×i1=

1×i1=

i1=

i

6 0
3 years ago
Find the sum of -3x2 -4x+3 and 2x2-x+3
Alika [10]

Answer:

Assuming the 2 at the ends are the exponents,

the answer is -x^2-5x+6 (x^2+5x-6)

from here, it cannot be simplified more

Explanation --

Just add everything by putting the terms in parentheses and adding them.

Keep simplifying by putting the like terms together and adding/subtracting them.

At the end, you come up with -x^2-5x+6, so multiply -1 on the whole thing (by putting parentheses on its sides. This makes x positive. But they are the same thing, but this step is just for simplifying the end product a little more.

Hope this helps!! Have a nice day!

5 0
3 years ago
Michelle’s mom is letting her paint her room pink. What formula would she use to figure out how much paint she needs.
postnew [5]
There’s no picture or anything so try again
6 0
3 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
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