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3 years ago

7

To convert a temperature in degrees celsius to degrees fahrenheit multiply the celsius temperature by 9/5 and then add 32° is th

e temperature in degrees celsius proportional to its equivalent temperature in degrees fahrenheit

1 answer:

babunello [35]3 years ago

7 0

A temperature in degree celcius is not equivalent to its equivalent temperature in degrees fahrenheit.

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The distribution of the number of transactions performed at a bank each day is approximately normal with mean 478 transactions a

jolli1 [7]

Answer:

e. 0.977

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 478, \sigma = 64$

Which of the following is closest to the proportion of daily transactions greater than 350?

This is 1 subtracted by the pvalue of Z when X = 350. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{350 - 478}{64}$

$Z = -2$

$Z = -2$ has a pvalue of 0.023

1 - 0.023 = 0.977

So the correct answer is:

e. 0.977

6 0

2 years ago

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side

lorasvet [3.4K]

Answer:

Step-by-step explanation:

1.

cot x sec⁴ x = cot x+2 tan x +tan³x

L.H.S = cot x sec⁴x

=cot x (sec²x)²

=cot x (1+tan²x)² [ ∵ sec²x=1+tan²x]

= cot x(1+ 2 tan²x +tan⁴x)

=cot x+ 2 cot x tan²x+cot x tan⁴x

=cot x +2 tan x + tan³x [ ∵cot x tan x $=\frac{ \textrm{tan x }}{\textrm{tan x}}$ =1]

=R.H.S

2.

(sin x)(tan x cos x - cot x cos x)=1-2 cos²x

L.H.S =(sin x)(tan x cos x - cot x cos x)

= sin x tan x cos x - sin x cot x cos x

$=\textrm{sin x cos x }\times\frac{\textrm{sin x}}{\textrm{cos x} } - \textrm{sinx}\times\frac{\textrm{cos x}}{\textrm{sin x}}\times \textrm{cos x}$

= sin²x -cos²x

=1-cos²x-cos²x

=1-2 cos²x

=R.H.S

3.

1+ sec²x sin²x =sec²x

L.H.S =1+ sec²x sin²x

= $1+\frac{{sin^2x}}{cos^2x}$ [ $\textrm{sec x}=\frac{1}{\textrm{cos x}}$ ]

=1+tan²x $[\frac{\textrm{sin x}}{\textrm{cos x}} = \textrm{tan x}]$

=sec²x

=R.H.S

4.

$\frac{\textrm{sinx}}{\textrm{1-cos x}} +\frac{\textrm{sinx}}{\textrm{1+cos x}} = \textrm{2 csc x}$

L.H.S= $\frac{\textrm{sinx}}{\textrm{1-cos x}} +\frac{\textrm{sinx}}{\textrm{1+cos x}}$

$=\frac{\textrm{sinx(1+cos x)+{\textrm{sinx(1-cos x)}}}}{\textrm{(1-cos x)\textrm{(1+cos x})}}$

$=\frac{\textrm{sinx+sin xcos x+{\textrm{sinx-sin xcos x}}}}{{(1-cos ^2x)}}$

$=\frac{\textrm{2sin x}}{sin^2 x}$

= 2 csc x

= R.H.S

5.

-tan²x + sec²x=1

L.H.S=-tan²x + sec²x

= sec²x-tan²x

= $\frac{1}{cos^2x} -\frac{sin^2x}{cos^2x}$

$=\frac{1- sin^2x}{cos^2x}$

$=\frac{cos^2x}{cos^2x}$

=1

8 0

3 years ago

List the 4-digit numbers that fit these clues

eimsori [14]

What are the clues fvfrce

6 0

3 years ago

Solve the system of linear equations using the elimination method: 2x + 2y = 16 -2x +3y = 14

horrorfan [7]

Answer:

Step-by-step explanation:

soulution:

given, 2x +2x = 16 <em>-</em> <em>2x</em> + 3y = 14

2x <em>+ 2x</em> = <em>-2y </em>+3y = 14 -16

4x = y = -2

y = 4x = -2

y = x = 4/-2

x = y = <u><em>-2 ans</em></u>

4 0

2 years ago

Applicants who wish to be admitted to a certain professional school in a large university are required to take a screening test

Elden [556K]

Answer:

Probability of eligible applicants who pass the exam is 0.665

probability of applicants who are ineligible but pass the exam 0.063

Step-by-step explanation:

Total percentage eligible applicants who pass the exam

$\textrm {89\% of 70\%}= \dfrac{89\times70}{100}=62.3\%$

Total ineligible applicants who pass the exam

$\textrm {14\% of 30\%}= \dfrac{14\times30}{100}=4.2\%$

All applicants who pass this exam 62.3% + 4.2% = 66.5%

Probability of applicants who pass the exam

$\textrm{probability }= \dfrac{66.5}{100}=0.665$

Out of 66.5% applicants who pass the exam , 4.2% applicants are ineligible

$\textrm{so the probability is}\dfrac{4.2}{66.5}=0.063$

Probability of applicants who pass the exam is 0.665

probability of applicants who are ineligible but pass the exam 0.063

6 0

2 years ago