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What is the value of a in the equation 3 a + b = 54, when b=9

slavikrds [6]

To solve this plug in 9 for b and solve for a like so...

3a + 9 = 54

9 must be combined with 54. To do this you must subtract 9 from both sides. This will cancel out 9 from the left and bring it over to the right

3a + (9 - 9) = 54 - 9

3a + 0 = 45

3a = 45

Now we must isolate the a. To do this you have to divide 3 to both sides. This will make 3 on the left side 1 and bring 3 over to the other side

3a/3 = 45/3

1a = 15

a = 15

Hope this helped!

~Just a girl in love with Shawn Mendes

6 0

3 years ago

An electric pole is 10m high if it's shadow is 10×√3 in length find the elevation ofthe sum

lianna [129]

First, Look at the attached picture as a reference.

Now, using Pythagoras theorem,

$AC^2=AB^2+BC^2 \\\\AC=\sqrt{AB^2+BC^2 }\\\\AC=\sqrt{10^2+(10\sqrt3)^2}\\\\ AC = \sqrt{100+300}\\\\AC=\sqrt{400}\\\\AC=20~in.$

Now, lets find the elevation of the sun.

 $tan\theta = \frac{AB}{BC} \\\\ tan\theta = \frac{10}{10\sqrt3} \\\\ \theta = tan^{-1} \left(\frac{10}{10\sqrt3}\right) \\\\ \boxed{\theta = 30 \°}$

∴ The elevation of the sun is 30°

6 0

3 years ago

Beer shelf life is a problem for brewers and distributors because when beer is stored at room temperature, its flavor deteriorat

Blizzard [7]

Answer:

a) 3.52

b) 0.002399

c) At α = .05 we reject the null hypothesis.

Step-by-step explanation:

We are given the following information:

Population mean, μ = 6 μg per liter

Sample size, n = 12

Alpha, α = 0.05

6.94, 7.50, 5.87, 7.92, 8.66, 7.78, 7.95, 6.35, 6.16, 6.28, 5.68, 7.02

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{84.11}{12} = 7.009$

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$SS.D = \sqrt{\frac{10.03}{11}} = 0.95$

First, we design the null and the alternate hypothesis

$H_{0}: \mu \leq 6\\H_A: \mu > 6$

We use One-tailed t test to perform this hypothesis.

a) Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n-1}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{7.009 - 6}{\frac{0.95}{\sqrt{11}} } = 3.52$

b) P-value for one tailed t-test at 11 degree of freedom and α = 0.05 is 0.002399

c) Since P-value < 0.05

The result is insignificant.

Thus, at α = .05 we reject the null hypothesis.

6 0

3 years ago

Original price: Discount: 30% Sale price: $206.50

REY [17]

The answer is $295. I also added my work below!

4 0

2 years ago

Arrivals of customers at a checkout counter follow a Poisson distribution. It is known that, during a given 30-minute period, on

Xelga [282]

Answer:

1/4

Step-by-step explanation:

Let X be the time of arrival measured from the beginning of the 30-minute period.

Since the number of arrivals is Poisson, the time of the arrival is equally likely in any subinterval of time of a given size in the 30 minutes, and thus X has a uniform distribution in (0,30) and P(X > 25) = $\int\limits^a_b {1/30} \, dx$

where a = 30 and b = 25

Solving, we obtain P(X > 25) = 1/6

similarly, P(X>10) = 2/3 by substituting b for 10

Hence, for conditional probability we have

P(X>25 | X>10) = P(X>25) / P(X>10) = (1/6) / (2/3) = 1/4

3 0

3 years ago