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Mice21 [21]
3 years ago
14

In the figure, a square is inside another bigger square.

Mathematics
2 answers:
evablogger [386]3 years ago
7 0

Answer:

Part 1) The length of the diagonal of the outside square is 9.9 units

Part 2) The length of the diagonal of the inside square is 7.1 units

Step-by-step explanation:

step 1

Find the length of the outside square

Let

x -----> the length of the outside square

c ----> the length of the inside square

we know that

x=a+b=4+3=7\ units

step 2

Find the length of the inside square

Applying the Pythagoras Theorem

c^{2}= a^{2}+b^{2}

substitute

c^{2}= 4^{2}+3^{2}

c^{2}=25

c=5\ units

step 3

Find the length of the diagonal of the outside square

To find the diagonal Apply the Pythagoras Theorem

Let

D -----> the length of the diagonal of the outside square

D^{2}= x^{2}+x^{2}

D^{2}= 7^{2}+7^{2}

D^{2}=98

D=9.9\ units

step 4

Find the length of the diagonal of the inside square

To find the diagonal Apply the Pythagoras Theorem

Let

d -----> the length of the diagonal of the inside square

d^{2}= c^{2}+c^{2}

d^{2}= 5^{2}+5^{2}

d^{2}=50

d=7.1\ units

coldgirl [10]3 years ago
3 0

Answer:

its 7 and 10 in the blanks im pretty sure.

Step-by-step explanation:

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Free_Kalibri [48]
4 1/3 

Okay, I know fractions are scary, but we can do this alright?

First, you see the whole number on the side? Get that first.

10- 5 = 5

now we have 5 left and a scary fraction.  Don't panic, let's do this.  We know 1 is 3/3.  1 can be anything as long as the number on top and the number at the bottom are the same then it would be one.

So 3/3 is one then we can subtract:

3/3 -2/3 .

We subtract the top number and leave the one at the bottom the same.

3-2 = 1

Then we have 1/3.

As you took one away from the 5, it becomes a 4 and you put back the left over, 1/3.

Then you answer will be :

4  1/3
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3 years ago
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A politician claims that a proposal for a new traffic law is broadly supported by both political parties and that a person from
Mashutka [201]

Answer:

Based on the 95% confidence interval for the difference in population proportion, there is convincing statistical evidence that he is correct

Step-by-step explanation:

The proportion from the sample of people from his party that support the law = 70%

The number the members of the politicians political party that support the law = 550 people

The proportion from the sample of people from the other party that support the law = 65%

The number the members of the politicians political party that support the law = 420 people

The confidence level of the test = 95%

The given confidence interval for the difference in proportion, C.I. = (-0.010, 0.110)

Given that the 95% confidence interval for the difference in population proportion ranges from -0.010, to 0.110, it is 95% certain that 0 is among the likely difference in proportion between the two populations and therefore, there is sufficient statistical evidence to suggest that there is no difference in the proportion of the members of either political that support the proposed new traffic law.

4 0
3 years ago
I need help with 6,7 pleaaseee
antiseptic1488 [7]
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5 0
3 years ago
Exhibit B: A restaurant has tracked the number of meals served at lunch over the last four weeks. The data shows little in terms
Mekhanik [1.2K]

Answer:

Option E is correct.

The expected number of meals expected to be served on Wednesday in week 5 = 74.2

Step-by-step Explanation:

We will use the data from the four weeks to obtain the fraction of total days that number of meals served at lunch on a Wednesday take and then subsequently check the expected number of meals served at lunch the next Wednesday.

Week

Day 1 2 3 4 | Total

Sunday 40 35 39 43 | 157

Monday 54 55 51 59 | 219

Tuesday 61 60 65 64 | 250

Wednesday 72 77 78 69 | 296

Thursday 89 80 81 79 | 329

Friday 91 90 99 95 | 375

Saturday 80 82 81 83 | 326

Total number of meals served at lunch over the 4 weeks = (157+219+250+296+329+375+326) = 1952

Total number of meals served at lunch on Wednesdays over the 4 weeks = 296

Fraction of total number of meals served at lunch over four weeks that were served on Wednesdays = (296/1952) = 0.1516393443

Total number of meals expected to be served in week 5 = 490

Number of meals expected to be served on Wednesday in week 5 = 0.1516393443 × 490 = 74.3

Checking the options,

74.3 ≈ 74.2

Hence, the expected number of meals expected to be served on Wednesday in week 5 = 74.2

Hope this Helps!!!

8 0
3 years ago
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