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3 years ago

Fully factorise 15ab - 35a

1 answer:

3 0

The answer is 5a (3b-7).
Step 1: find the common factor. since both 15ab and 35a has a common "a" and a factor of "5", you factorise them out.
Step 2: place the common factor at the front and divide the equation from the common factor.
5a (3b-7) = 15ab - 35a

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What is the smallest positive degree angle measure equivalent to sin^-1 (0.391)?

natta225 [31]

Given:

The value is:

$\sin^{-1}(0.391)$

To find:

The smallest positive degree angle measure equivalent to $\sin^{-1}(0.391)$ .

Solution:

We have,

$\sin^{-1}(0.391)$

Using the scientific calculator, we get

$\sin^{-1}(0.391)=23.0167366398^\circ$

$\sin^{-1}(0.391)\approx 23^\circ$

Therefore, the smallest positive degree angle measure equivalent to $\sin^{-1}(0.391)$ is 23 degrees.

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3 years ago

In a sample of eight whitefish caught in Yellowknife Bay, the mean arsenic concentration in the liver was 0.32 mg/kg, with a sta

Advocard [28]

Answer:

The 95% confidence interval for the concentration in whitefish found in Yellowknife Bay is (0.2698 mg/kg, 0.3702 mg/kg).

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 8 - 1 = 7

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 7 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 2.3246

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.3646\frac{0.06}{\sqrt{8}} = 0.0502$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 0.32 - 0.0502 = 0.2698 mg/kg

The upper end of the interval is the sample mean added to M. So it is 0.32 + 0.0502 = 0.3702 mg/kg

The 95% confidence interval for the concentration in whitefish found in Yellowknife Bay is (0.2698 mg/kg, 0.3702 mg/kg).

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2 years ago

Need help with question number 2

USPshnik [31]

Answer:

BC=20

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Step-by-step explanation:

To find DAB you subtract 180 from 114

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3 years ago

The half-life of caffeine in a healthy adult is 4.8 hours. Jeremiah drinks 18 ounces of caffeinated

statuscvo [17]

We want to see how long will take a healthy adult to reduce the caffeine in his body to a 60%. We will find that the answer is 3.55 hours.

We know that the half-life of caffeine is 4.8 hours, this means that for a given initial quantity of coffee A, after 4.8 hours that quantity reduces to A/2.

So we can define the proportion of coffee that Jeremiah has in his body as:

P(t) = 1*e^{k*t}

Such that:

P(4.8 h) = 0.5 = 1*e^{k*4.8}

Then, if we apply the natural logarithm we get:

Ln(0.5) = Ln(e^{k*4.8})

Ln(0.5) = k*4.8

Ln(0.5)/4.8 = k = -0.144

Then the equation is:

P(t) = 1*e^{-0.144*t}

Now we want to find the time such that the caffeine in his body is the 60% of what he drank that morning, then we must solve:

P(t) = 0.6 = 1*e^{-0.144*t}

Again, we use the natural logarithm:

Ln(0.6) = Ln(e^{-0.144*t})

Ln(0.6) = -0.144*t

Ln(0.6)/-0.144 = t = 3.55

So after 3.55 hours only the 60% of the coffee that he drank that morning will still be in his body.

If you want to learn more, you can read:

brainly.com/question/19599469

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2 years ago