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2 years ago

PLEASE HELPPppp

Mathematics

1 answer:

sasho [114]2 years ago

5 0

Answer: B

Hope this helps :)

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Parallelogram ABCD has diagonal segments AC and BD that intersect at E. if BE is ( x + 8) and ED is (5 x), find the length of BD

Virty [35]

Answer:

BD = 20 units

Step-by-step explanation:

Given that,

Parallelogram ABCD has diagonal segments AC and BD that intersect at E.

BE = (x+8) and ED = 5x

We need to find the length of BD.

BD = BE + ED ...(1)

We know that the diagonals of a parallelogram bisects each other. It means,

BE = ED

x+8 = 5x

8 = 5x - x

8 = 4x

x = 2

BE = (x+8)

= (2+8)

= 10

ED = 5x

= 5(2)

= 10

Put the values in equation (1).

BD = 10 + 10

= 20 units

Hence, the length of BD is 20 units.

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2 years ago

Nora ordered a set of purple and green pins. She received 50 pins, and 70% of them were

Paladinen [302]

Answer:

Nora received <u>35</u> purple pins.

Step-by-step explanation:

If Nora received 50 pins, and 70% of the pins were purple, to calculate the number of purple pins Nora received, we must find 70% of 50.

To find 70% of 50, we can multiply 50 by the decimal form of 70%:

$50 * .7=35$

Therefore, we can conclude that since 70% of the 50 pins Nora received were purple, and 70% of 50 is 35, Nora received 35 purple pins.

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2 years ago

7. A boutique is offering a 25% discount on purchase made in cash. Calculate the cash price Janelle paid for articles totaled to

igor_vitrenko [27]

Answer:

$348.75

Step-by-step explanation:

$465 * 25%= 116.25

465-116.25 = $348.75

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2 years ago

How to write 10,010,468 in word form

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Ten million ten thousand four hundred sixty eight

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3 years ago

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Seven students volunteered for a comparison of study guides for an advanced course in mathematics. They were randomly assigned,

andriy [413]

Answer:

Step-by-step explanation:

From the given question; we can use the R software to program the combination function that generates all the combinations.

options(digits =2(

scores<- c(68,77,82,85,53,64,71)

groupA <- combn(scores,4)

groupB <- apply(groupA,2, function(x) scores[! (scores %in% x) ] )

colnames(groupA) <- colnames(groupB) <- paste("G", 1:35, sep"")

The accompanying 35 groupings (G1 to G35) contain all potential ways these understudies can be randomized under the null hypothesis

Group A

$\text{G1 \ G2 \ G3 \ G4 \ G5 \ G6 \ G7 \ G8 \ G9 \ G10\ G11\ G12 \ G13 \ G14}$

$\text{68 \ \ 68 \ \ 68 \ \ 68 \ \ 68 \ \ 68 \ \ 68 \ \ 68 \ \ 68 \ \ 68 \ \ 68 \ \ 68 \ \ 68 \ \ 68}$

$\text{77 \ \ 77 \ \ 77 \ \ 77 \ \ 77 \ \ 77 \ \ 77 \ \ 77 \ \ 77 \ \ 77 \ \ 82 \ \ 82 \ \ 82 \ \ 82}$

$\text{82 \ \ 82 \ \ 82 \ \ 82 \ \ 85 \ \ 85 \ \ 85 \ \ 53 \ \ 53 \ \ 64 \ \ 85 \ \ 85 \ \ 85 \ \ 53}$

$\text{85\ \ 53 \ \ 64 \ \ 71 \ \ 53 \ \ 64\ \ 71\ \ 64 \ \ 71 \ \ 71 \ \ \ 53 \ \ \ 64 \ \ 71 \ \ 64}$

$\text{G15 G16 G17 G18 G19 G20 G21 G22 \ G23 \ G24 \ G25 \ G26 \ G27}$

$\text{68 \ \ \ 68 \ \ \ 68 \ \ \ 68 \ \ \ 68 \ \ \ 68 \ \ \ 77 \ \ \ 77 \ \ \ 77 \ \ \ 77 \ \ \ 77 \ \ \ 77 \ \ \ 77}$

$\text{82 \ \ \ 82 \ \ \ 85 \ \ \ 85 \ \ \ 85 \ \ \ 53 \ \ \ 82 \ \ \ 82 \ \ \ 82 \ \ \ 82 \ \ \ 82 \ \ \ 82 \ \ \ 85}$

$\text{53 \ \ \ 64 \ \ \ 53 \ \ \ 53 \ \ \ 64 \ \ \ 64 \ \ \ 85 \ \ \ 85 \ \ \ 85 \ \ \ 53 \ \ \ 53 \ \ \ 64 \ \ \ 53}$

$\text{71\ \ \ \ 71\ \ \ \ 64\ \ \ \ \ 71\ \ \ \ 71\ \ \ \ 71\ \ \ \ 53\ \ \ \ 64\ \ \ \ 71\ \ \ 64\ \ \ \ 71\ \ \ \ 71\ \ \ \ 64}$

$\text{G28 G29 G30 G31 G32 G33 G34 \ G35} \\ \\ 77 \ \ \ \ 77 \ \ 77 \ \ \ \ 82\ \ \ \ 82 \ \ \ 82 \ \ \ 82 \ \ \ \ \ 85 \\ \\ 85 \ \ \ 85 \ \ \ 53 \ \ \ \ 85 \ \ \ 85 \ \ \ 85 \ \ \ \ 53 \ \ \ 53 \\ \\ 53 \ \ \ 64 \ \ 64 \ \ \ 53 \ \ \ \ 53 \ \ \ 64 \ \ \ \ 64\ \ \ 64 \\ \\ 71 \ \ 71 \ \ \ 71 \ \ 64 \ \ \ 71 \ \ \ \ 71 \ \ \ \ 71 \ \ \ \ 71$

Group B

$\tet{53 \ \ 85 \ \ \ \ 85 \ \ \ \ 85\ \ \ \ 82 \ \ \ \ 82\ \ \ \ 82 \ \ \ \ 82\ \ \ \ 82 \ \ \ \ 82 \ \ \ \ 77\ \ \ \ 77\ \ \ \ 77\ \ \ \ 77\ \ \ \ 77}$

$\text{64 \ \ \ 64 \ \ \ 53 \ \ \ 53 \ \ \ 64 \ \ \ 53 \ \ \ 53 \ \ \ 85 \ \ \ 85 \ \ \ 85 \ \ \ 64 \ \ \ 53 \ \ \ 53 \ \ \ 85 \ \ \ 85}$

$\text{71 \ \ \ 71 \ \ \ 71 \ \ \ 64 \ \ \ 71 \ \ \ 71 \ \ \ 64 \ \ \ 71 \ \ \ 64 \ \ \ 53 \ \ \ 71 \ \ \ 71 \ \ \ 64 \ \ \ 71 \ \ \ 64}$

$\text{G16 \ G17 \ G18 \ G19 \ G20 \ G21 \ G22\ \ G23\ \ G24\ \ G25 \ \ G26 \ \ G27\ \ G28}$

$\text{77\ \ \ \ 77\ \ \ \ 77\ \ \ \ \ 77\ \ \ \ \ 77\ \ \ \ \ 68\ \ \ \ 68\ \ \ \ 68\ \ \ \ 68\ \ \ \ 68\ \ \ \ \ 68\ \ \ \ \ 68\ \ \ \ \ 68}$

$\text{85 \ \ \ \ 82\ \ \ \ 82 \ \ \ \ 82 \ \ \ \ 82 \ \ \ \ 64 \ \ \ \ 53 \ \ \ \ 53 \ \ \ \ 85 \ \ \ \ 85\ \ \ \ 85 \ \ \ \ 82\ \ \ \ 82}$

$\text{53\ \ \ \ 71\ \ \ \ 64\ \ \ \ 53\ \ \ \ 85\ \ \ \ 71\ \ \ \ 71\ \ \ \ 64\ \ \ \ 71\ \ \ \ 64\ \ \ \ 53\ \ \ \ 71\ \ \ \ 64}$

$\text{ G29 \ G30\ G31 \ G32 \ G33 \ G34 \ G35} \\ \\ \text{68 \ \ \ 68 \ \ \ 68 \ \ \ \ 68 \ \ \ \ 68 \ \ \ 68 \ \ \ \ \ 68} \\ \\ \text{82 \ \ \ 82 \ \ \ \ 77 \ \ \ \ 77 \ \ \ \ 77 \ \ \ \ 77 \ \ \ \ 77} \\ \\ \text{53 \ \ \ 85 \ \ \ \ 71 \ \ \ \ 64 \ \ \ \ 53\ \ \ \ 85 \ \ \ \ 82} \\ \\$

The accompanying data below computes the distinctions for each group:

$difference$

$\text{G1 G2 G3 G4 G5 G6 G7 G8 G9 \ G10 G11 G12 G13 G14 G15} \\ \\ \text{15 -3.3 3.1 7.2 -1.6 4.8 8.9 -14 -9.8 -3.3 1.3 \ 7.8 \ 12 \ -11 \ \ -6.8}$

$\text{G16 G17 G18 G19 G20 G21 G22 G23 G24 G25 G26 G27 G28} \\ \\ \text{-0.42 \ -9.2\ -5.1\ 1.3 \ -17\ \ 6.6 \ \ 13 \ 17 \ \ -5.7\ -1.6 \ \ 4.8 \ \ -3.9 \ \ 0.17}$

$\text{ G29\ \ G30 \ G31 \ G32 \ \ G33 \ G34 \ \ G35} \\ \\ \text{6.6 \ \ \ -12 \ \ \ -1 \ \ \ 3.1 \ \ \ 9.5 \ \ \ -9.2 \ \ \ -7.4}$

The two-sided p-value is the extent of contrasts between test midpoints as large or bigger in supreme value than the primary group. The cat function makes the outcomes simpler to peruse.

p <- sum (aba(difference)>=difference[1])/35

cat(p)

= 0.086

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2 years ago