The value of y varies directly with x, and y=28 when x=7. Find y when x=4 by first finding k.

1 answer:

7 0

Direct proportionality is the type of proportionality where an increase in one quantity or variable causes a corresponding increase in another quantity.
Therefore; y varies directly as x
Hence; y α x
y =k x where k is a constant
if y= 28 and x = 7 , then k = 28/4 =4
hence, y = 4 k
when x = 4, then y= 16

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A data set includes 103 body temperatures of healthy adult humans having a mean of 98.9degreesf and a standard deviation of 0.67

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Answer:

Step-by-step explanation:

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Margin of error = z × s/√n

Where

s = sample standard deviation = 0.67

n = number of samples = 103

From the information given, the population standard deviation is unknown hence, we would use the t distribution to find the z score

In order to use the t distribution, we would determine the degree of freedom, df for the sample.

df = n - 1 = 103 - 1 = 102

Since confidence level = 99% = 0.99, α = 1 - CL = 1 – 0.99 = 0.01

α/2 = 0.01/2 = 0.005

the area to the right of z0.005 is 0.005 and the area to the left of z0.005 is 1 - 0.005 = 0.995

Looking at the t distribution table,

z = 2.6249

Margin of error = 2.6249 × 0.67/√103

= 0.173

Confidence interval = 98.6 ± 0.173

This suggests that the mean body temperature could very possibly be

98.6degrees°F.

3 0

3 years ago

What is the coefficient of the x^5y^5- term in the biomial expansion of (2x-3y)^10

jasenka [17]

<u>Answer:</u>

The coefficient of $x^{5} \times y^{5} \text { is }=\left(252 \times 2^{5} \times(-3)^{5}\right)=252 \times 32 \times 243=1959552$

<u>Solution: </u>

The given expression is $(2 x-3 y)^{10}$

As per binomial theorem, we know,

$(x+y)^{n}=\sum n C_{k} x^{n-k} y^{k}$

Now here a = 2x, b = (- 3y) and n = 10 and k = 0,1,2,….10

Now $x^{5}\times y^5$ will be the 6 term where k =5

Now, $\mathrm{T}_{6}=10 \mathrm{C}_{5} \times(2 \mathrm{x})^{(10-5)} \times(-3 \mathrm{y})^{5}=10 \mathrm{C}_{5} 2^{5} \times \mathrm{x}^{5} \times(-3)^{5} \times \mathrm{y}^{5}$

So, the coefficient of $x^{5} \times y^{5} \text { is }=10\left(5 \times 2^{5} \times(-3)^{5}\right$ .

$10 \mathrm{C}_{5}=\frac{10 !}{5 ! \times(10-5) !}=\frac{10 !}{5 !+5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}=\left(\frac{30240}{120}\right)=252$