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miskamm [114]
3 years ago
14

if the HCF of 408 and 1032 is expressible in the form of 1032 into 2 + 408 into P then find the value of p​

Mathematics
1 answer:
alexandr1967 [171]3 years ago
3 0

Answer:

below.

Step-by-step explanation:

408 = 2*2*2*3*17

1032 = 2*2*2*3*43

So the HCF is 2*2*2*3 = 24

24 = 2/1032 + p/408

24 - 2/1032 = p/408

p = (24 - 2/1032) * 408

p = 9791.21 to nearest hundredth.

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Assume that all conditions are met. The mean of the differences was 1. 33 and the standard deviation of those differences was 2.
Maurinko [17]

The test statistic for this procedure is using two sample T-test is 0.458.

Definition of Two Sample T-test

A variation of the Student's t-test used to determine if two sample means are substantially different is known as the two sample T-test (also known as Welch's t-test, Welch's adjusted T, or unequal variances t-test). The test's degrees of freedom have been changed, which generally boosts the test's power for samples with unequal variance.

T-Test Statistic for The Procedure

It is given that mean of the differences = 1.33

And, the standard deviation of the differences, standard error = 2.90

The test static formula is given as,

t = (estimate - hypothized value) / standard error

Here,  t is the t- test statistic, and estimate is the mean of the differences.

Since it is given that all conditions are met, the hypothized value can be taken as 0

Substituting substituting estimate = 1.33 and standard error = 2.90 in the t-test statistic formula, we get,

t = (1.33-0)/2.90

or t = 0.458

learn more about t-test here:

brainly.com/question/14128303

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8 0
2 years ago
164 is 55% of what number?<br><br> Please somebody help me
monitta
This is how you solve it; but first of all convert the percent into a decimal;55%= 0.55

0.55x = 164 (divide by 0.55 on both sides)
x = 164/0.55 
x = 298.18 (rounded to the nearest hundredth)
so 164 is 55% of 298.18(rounded)
8 0
3 years ago
PLEASE HELP ILL GIVE BRAINLIEST!!
bekas [8.4K]

24 inches hope this helps

7 0
3 years ago
Read 2 more answers
In ΔABC, m∠B = m∠C. The angle bisector of ∠B meets AC at point H and the angle bisector of ∠C meets AB at point K. Prove that BH
solniwko [45]

Answer:

See explanation

Step-by-step explanation:

In ΔABC, m∠B = m∠C.

BH is angle B bisector, then by definition of angle bisector

∠CBH ≅ ∠HBK

m∠CBH = m∠HBK = 1/2m∠B

CK is angle C bisector, then by definition of angle bisector

∠BCK ≅ ∠KCH

m∠BCK = m∠KCH = 1/2m∠C

Since m∠B = m∠C, then

m∠CBH = m∠HBK = 1/2m∠B = 1/2m∠C = m∠BCK = m∠KCH   (*)

Consider triangles CBH and BCK. In these triangles,

  • ∠CBH ≅ ∠BCK (from equality (*));
  • ∠HCB ≅ ∠KBC, because m∠B = m∠C;
  • BC ≅CB by reflexive property.

So, triangles CBH and BCK are congruent by ASA postulate.

Congruent triangles have congruent corresponding sides, hence

BH ≅ CK.

5 0
3 years ago
A point is reflected across the y axis the new point is located at -4.25,-1.75 write the order pair that represents the original
Naya [18.7K]

Answer:

(4.25, - 1.75)

Step-by-step explanation:

Under a reflection in the y- axis

a point (x, y ) → (- x, y ), thus

(- 4.25, - 1.75 ) → (4.25, - 1.75 ) ← original point



5 0
3 years ago
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