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Svetlanka [38]
3 years ago
7

Can someone help me figure this out

Mathematics
2 answers:
Ymorist [56]3 years ago
3 0
There is no question or photo attached.
12345 [234]3 years ago
3 0
5 + 1.5(2d - 1) = 0.5
5 + 3d - 1.5 = 0.5
3.5 + 3d = 0.5
3d = -3
d = -1

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Which polynomial correctly combines the like terms and expresses the given polynomial in standard form? 8mn5 – 2m6 + 5m2n4 – m3n
MaRussiya [10]

Answer:

-6m^{6}+n^{6}+7mn^{5}+14m^{2}n^{4}-5m^{3}n^{3}

Step-by-step explanation:

given polynomial is:

8mn^{5}-2m^{6}+5m^{2}n^{4}-m^{3}n^{3}+n^{6}-4m^{6}+9m^{2}n^{4}- mn^{5}-4m^{3}n^{3}

write similar terms together,

-2m^{6}-4m^{6}+n^{6}+8mn^{5}-mn^{5}+5m^{2}n^{4}+9m^{2}n^{4}-m^{3}n^{3}-4m^{3}n^{3}

-6m^{6}+n^{6}+7mn^{5}+14m^{2}n^{4}-5m^{3}n^{3}

Hence, polynomial in standard form -6m^{6}+n^{6}+7mn^{5}+14m^{2}n^{4}-5m^{3}n^{3}

8 0
3 years ago
What is two thirds takeaway a half?
Karo-lina-s [1.5K]
\displaystyle  \frac{2}{3} - \frac{1}{2} = \frac{4}{6} - \frac{3}{6} = \frac{1}{6}
8 0
3 years ago
Which of the following congruence statements does not prove that two triangles are congruent?
steposvetlana [31]
There are three postulates that can be used to prove two that two triangles are congruent. 

ASA or Angle-Side-Angle: If two corresponding interior angles of two triangles and the included sides are congruent, the triangles are congruent.

SAS or Side-Angle-Side: If two corresponding sides of two triangles and the included angles are congruent, the triangles are congruent.

SSS or Side-Side-Side: If three corresponding sides of two triangles are congruent, the triangles are congruent.

SSA is not a congruence statement. 
  
6 0
3 years ago
A state park has two pools. The olympic size pool holds 8.12 x 105 gallons of water and the smaller pool holds 5.27 x 105 gallon
mr_godi [17]
First you need to work out what (8.12×105=?) Then (5.27×105=?) And add them together so (8.12×105)+(5.27×105)=the answer to your question
6 0
3 years ago
Read 2 more answers
Question 3 A study was performed to determine whether men and women differ in their repeatability in assembling components on pr
Fed [463]

Answer:

There is no sufficient evidence to support the claim that men and women differ in repeatability for this assembly task

Step-by-step explanation:

Given

Let subscript 1 represent men and 2 represent women, respectively.

n_1 = 25

n_2 = 21

s_1 = 0.98

s_2 = 1.02

\alpha = 0.02

Required

Determine if here is enough evidence

First, we need to state the hypotheses

H_o: \sigma_1^2 = \sigma_2^2

H_1: \sigma_1^2 \ne \sigma_2^2

Next, calculate the test statistic using:

F = \frac{s_1^2}{s_2^2}

F = \frac{0.98^2}{1.02^2}

F = 0.923

Calculate the rejection region;

But first, calculate the degrees of freedom

df_1 =n_1 - 1

df_1 =25 - 1

df_1 =24

df_2 = n_2 -1

df_2 = 21 - 1

df_2 = 20

Using the F Distribution:  table

c = \frac{\alpha}{2}

c = \frac{0.02}{2}

c = 0.01

At 0.01 level (check row 20 and column 24), the critical value is:

<em></em>f_{0.01,24,20} = 2.86<em> --- the upper bound</em>

At 0.01 level (check row 24 and column 20), the critical value is:

f_{0.01,20,24} = 2.74

Calculate the inverse F distribution.

f_{0.99,20,24} = \frac{1}{f_{0.01,20,24}} = \frac{1}{2.74} =0.365 ---- the lower bound

The rejection region is then represented as:

0.365 < Test\ Statistic < 2.86

If the test statistic falls within this region, then the null hypothesis is rejected

F = 0.923 --- Test Statistic

0.365 < 0.923 < 2.86

<em>The above inequality is true; so, the null hypothesis is rejected.</em>

<em>This implies that, there is no sufficient evidence.</em>

7 0
3 years ago
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