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Dimas [21]
3 years ago
13

6= -s + 77 Given the above equation, what is the value of 1 + 5(77 - s)?

Mathematics
2 answers:
sveta [45]3 years ago
7 0

Answer:

1 + 5(77 - s) = 31

Step-by-step explanation:

Solve for s:

-s+77=6\qquad\text{subtract 77 from both sides}\\\\-s=-71\qquad\text{change the signs}\\\\s=71

Put the value of <em>s</em> to the expression 1 + 5(77 -s):

1+5(77-71)

Use PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

=1+5(6)=1+30=31

marysya [2.9K]3 years ago
4 0

Answer:

\boxed{\bold{31}}

Explanation:

Solve \bold{6= -s + 77}

--------------------------

Add 's' To Both Sides

= \bold{6+s=-s+77+s}

Simplify

= \bold{6+s=77}

Subtract 6 From Both Sides

= \bold{6+s-6=77-6}

Simplify

= \bold{s \ = \ 71}

Insert '71' Into 1 + 5 (77 - s)

-------------------------------

1 + 5 (77 - 71)

Solve:

77 - 71 = 6

= 1 + 5 · 6

5 · 6 = 30

= 1 + 30

= 31

Mordancy.

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In Which Quadrant is this true
34kurt

Given:

\sin \theta

\tan \theta

To find:

The quadrant in which \theta lie.

Solution:

Quadrant concept:

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II, only \sin\theta and \csc\theta are positive.

In Quadrant III, only \tan\theta and \cot\theta are positive.

In Quadrant IV, only \cos\theta and \sec\theta are positive.

We have,

\sin \theta

\tan \theta

Here, \sin\theta is negative and \tan\theta is also negative. It is possible, if \theta lies in the Quadrant IV.

Therefore, the correct option is D.

5 0
3 years ago
graph the equation by plotting three points if all three are correct the line will appear -4y=x minus 18
Alexandra [31]
Hope this image helps! Sorry if it's hard to read, I had to use the mouse to write it.

The points are: 
(-6,6)
(-2,5)
(6,3)

4 0
3 years ago
Solve for m. 5m+7/2 = −2m+5/2<br><br> a. -1/7<br> b. 6/7<br> c. 7<br> d. -1
Ainat [17]

5m+\dfrac{7}{2}=-2m+\dfrac{5}{2}\qquad|\text{add 2m to both sides}\\\\7m+\dfrac{7}{2}=\dfrac{5}{2}\qquad|\text{subtract}\ \dfrac{7}{2}\ \text{from both sides}\\\\7m=\dfrac{5-7}{2}\\\\7m=\dfrac{-2}{2}\\\\7m=-1\qquad|\text{divide both sides by 7}\\\\\boxed{m=-\dfrac{1}{7}}

8 0
3 years ago
Each day a commuter takes a bus to work, the transportation system has a phone app that tells her what time the bus will arrive.
Paraphin [41]

Answer:

Step-by-step explanation:

Hello!

The commuter is interested in testing if the arrival time showed in the phone app is the same, or similar to the arrival time in real life.

For this, she piked 24 random times for 6 weeks and measured the difference between the actual arrival time and the app estimated time.

The established variable has a normal distribution with a standard deviation of σ= 2 min.

From the taken sample an average time difference of X[bar]= 0.77 was obtained.

If the app is correct, the true mean should be around cero, symbolically: μ=0

a. The hypotheses are:

H₀:μ=0

H₁:μ≠0

b. This test is a one-sample test for the population mean. To be able to do it you need the study variable to be at least normal. It is informed in the test that the population is normal, so the variable "difference between actual arrival time and estimated arrival time" has a normal distribution and the population variance is known, so you can conduct the test using the standard normal distribution.

c.

Z_{H_0}= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } }

Z_{H_0}= \frac{0.77-0}{\frac{2}{\sqrt{24} } }= 1.89

d. This hypothesis test is two-tailed and so is the p-value.

p-value: P(Z≤-1.89)+P(Z≥1.89)= P(Z≤-1.89)+(1 - P(Z≤1.89))= 0.029 + (1 - 0.971)= 0.058

e. 90% CI

Z_{1-\alpha /2}= Z_{0.95}= 1.645

X[bar] ± Z_{1-\alpha /2}* (\frac{Sigma}{\sqrt{n} } )

0.77 ± 1.645 * (\frac{2}{\sqrt{24} } )

[0.098;1.442]

I hope this helps!

4 0
3 years ago
Please answer the question below, answers would be appreciated. And brainliest will be given out.
Andre45 [30]
4:12
Ana:
4
8
16

Frank:
12
24
48
8 0
3 years ago
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