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Marta_Voda [28]
3 years ago
15

Suppose that y is directly proportional to x, and y = 150 when x = 15. What is the constant of proportionality?

Mathematics
1 answer:
galina1969 [7]3 years ago
3 0

Answer:  The required constant of proportionality is 10.

Step-by-step explanation:  Given that y is directly proportional to x, and y = 150 when x = 15.

We are to find the constant of proportionality.

According to the given information, we can write

y\propto x\\\\\Rightarrow y=kx~~~~~~~~~~~~~~~~[\textup{where k is the constsnt of propotionality}].

When y = 150 and x = 15, we get from the above equation that

y=kx\\\\\Rightarrow 150=k\times15\\\\\Rightarrow k=\dfrac{150}{15}\\\\\Rightarrow k=10.

Thus, the required constant of proportionality is 10.

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How do you solve this
lys-0071 [83]
2x^2 + 14x = -12

Step 1: Subtract -12 from both sides.

2x^2 + 14x - (-12) = -12 - (-12)

2x^2 + 14x + 12 = 0

Step 2: Factor the left side of the equation.

2(x + 1)(x + 6) = 0

Step 3: Set Factors equal to 0.

x + 1 = 0 or x + 6 = 0

x = -1 or x = -6

Answer:
x = -1 or x = -6

Hope this helps. : )
5 0
4 years ago
for the data values below construct a 95 confidence interval if the sample mean is known to be 12898 and the standard deviation
Volgvan

Answer:

A 95% confidence interval for the population mean is [3315.13, 22480.87] .

Step-by-step explanation:

We are given that for quality control purposes, we collect a sample of 200 items and find 24 defective items.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                             P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~  t_n_-_1

where, \bar X = sample proportion of defective items = 12,898

             s = sample standard deviation = 7,719

            n = sample size = 5

             \mu = population mean

<em> Here for constructing a 95% confidence interval we have used a One-sample t-test statistics as we don't know about population standard deviation. </em>

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ; </u>

P(-2.776 < t_4 < 2.776) = 0.95  {As the critical value of t at 4 degrees of

                                               freedom are -2.776 & 2.776 with P = 2.5%}  

P(-2.776 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.776) = 0.95

P( -2.776 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.776 \times {\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-2.776 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.776 \times {\frac{s}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X-2.776 \times {\frac{s}{\sqrt{n} } } , \bar X+2.776 \times {\frac{s}{\sqrt{n} } } ]

                                = [ 12,898-2.776 \times {\frac{7,719}{\sqrt{5} } } , 12,898+2.776 \times {\frac{7,719}{\sqrt{5} } } ]

                               = [3315.13, 22480.87]

Therefore, a 95% confidence interval for the population mean is [3315.13, 22480.87] .

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(ANSWER ASAP) Solve the inequality.​
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Step-by-step explanation:

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