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MariettaO [177]
3 years ago
15

Use the information given to write the equation in point-slope form. slope

Mathematics
1 answer:
andrey2020 [161]3 years ago
5 0

Answer:

y + 1 = 5(x -3)

Step-by-step explanation:

y + 1 = 5(x -3)

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Solve for the exterior angle NEEDD HELP FAST
Sauron [17]

Answer:

x=76 degrees

Step-by-step explanation:

x+59=135

    -59   -59

x=76

i am 90% positive this is how you do it. Hope this helps! Happy holidays!!

6 0
3 years ago
Evaluate the expression: <br> 2^{4} ×4-\frac{2}{8}[<br> Plz answer
NISA [10]

Answer:

63.75

Step-by-step explanation:

2^4=16

16x4=64

64-2/8=63.75

because 2/8 can be reduced to 1/4 and 1/4 in decimal form is 0.25.

3 0
3 years ago
Read 2 more answers
A local deli sells 4-inch sub sandwiches for $2.95. It has decided to sell a “family sub” that is 50 inches long. How much shoul
lilavasa [31]
You take 50/4 which equals 12.5

so now you say “well one sub is 2.95”
take that and multiply it by 12

12 x 2.95 = 35.4

then you take 2.95 and divide it by 2 that is 1.48

so now you take 35.4 + 1.48

the cost should be 36.88
5 0
3 years ago
What is the area of this.
Ratling [72]
8x16=? 6x12=? Then add both answers ?+?=?
8 0
3 years ago
Use the following prompt to answer the next 6 questions. Suppose we want to test the color distribution claim on the M&amp;M’s w
vitfil [10]

Answer:

The claim on the M&M’s website is not true.

Step-by-step explanation:

A Chi-square test for goodness of fit will be used in this case.

The hypothesis can be defined as:

<em>H</em>₀: The observed frequencies are same as the expected frequencies.

<em>Hₐ</em>: The observed frequencies are not same as the expected frequencies.

The test statistic is given as follows:

\chi^{2}=\sum\limits^{n}_{i=1}{\frac{(O_{i}-E_{i})^{2}}{E_{i}}}

Here,

O_{i} = Observed frequencies

E_{i}=N\times p_{i} = Expected frequency.

The chi-square test statistic value is, 14.433.

The degrees of freedom is:

df = <em>k</em> - 1 = 6 - 1 = 5

Compute the <em>p</em>-value as follows:

p-value=P(\chi^{2}_{k-1} >14.433) =P(\chi^{2}_{5} >14.433) =0.013

*Use a Chi-square table.

The significance level is, <em>α</em> = 0.05.

p-value = 0.013 < <em>α</em> = 0.05.

So, the null hypothesis will be rejected at 5% significance level.

Thus, concluding that the claim on the M&M’s website is not true.

6 0
3 years ago
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