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2 years ago

Which digit is in the hundredths place in the number 742.891?

Mathematics

2 answers:

shtirl [24]2 years ago

7 0

Answer:

Step-by-step explanation:

fgiga [73]2 years ago

5 0

Answer:

Step-by-step explanation:

The hundredths place is the 2 one from the decimal point, which is 9.

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Which of the following is a solution to 2cos x + 1 = 0? 60° 120° 210° 300°

klasskru [66]

Rewrite <span>2cos x + 1 = 0 as:

2 cos x = -1, and then cos x = -1/2

x must be in Quadrant II or Quadrant III, since the adj. side is negative.

Note that the angle 120 has adj. side -1 and hyp 2. So 120 degrees is one solution.

Now what about a possible 2nd solution, to be found in Quadrant III? That would be -120 degrees, which has the same terminal line as does 240 degrees.

No soap.

So, the solution is 120 degrees.</span>

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3 years ago

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The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first deter

VMariaS [17]

Answer:

He must survey 123 adults.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

Assume that a recent survey suggests that about 87% of adults have heard of the brand.

This means that $\pi = 0.87$

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

How many adults must he survey in order to be 90% confident that his estimate is within five percentage points of the true population percentage?

This is n for which M = 0.05. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.05 = 1.645\sqrt{\frac{0.87*0.13}{n}}$

$0.05\sqrt{n} = 1.645\sqrt{0.87*0.13}$

$\sqrt{n} = \frac{1.645\sqrt{0.87*0.13}}{0.05}$

$(\sqrt{n})^2 = (\frac{1.645\sqrt{0.87*0.13}}{0.05})^2$

$n = 122.4$

Rounding up:

He must survey 123 adults.

6 0

2 years ago

Is this a function or not a function

notka56 [123]

this is a function becuase it passes the verticle line test (no 2 points are above each other).

8 0

2 years ago

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Two brothers decide that a 21-hour task is to be divided so thst the older brother works twice as long as the younger brother. H

lyudmila [28]

Answer:

14 hours

Step-by-step explanation:

Let's call the task 3x

The older bro does twice as younger so it would be 2x long and younger brother work x long

Since the totsl of work is 3x = 21 x woul be 21÷3 = 7

the older brother must work 2 × 7 = 14 hours long

3 0

2 years ago

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A tobacco company claims that its best-selling cigarettes contain at most 40 mg of nicotine. The average nicotine content from a

tatuchka [14]

Answer:

$t=\frac{42.6-40}{\frac{3.7}{\sqrt{15}}}=2.722$

$df=n-1=15-1=14$

$p_v =P(t_{(14)}>2.722)=0.0083$

We see that the p value is lower than the significance level given of 0.01 so then we can conclude that the true mean for the nicotine content is significantly higher than 40 mg

Step-by-step explanation:

Information provided

$\bar X=42.6$ represent the average nicotine content

$s=3.7$ represent the sample standard deviation

$n=15$ sample size

$\mu_o =40$ represent the value to check

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We want to verify if the nicotine content of the cigarettes exceeds 40 mg , the system of hypothesis are:

Null hypothesis: $\mu \leq 40$

Alternative hypothesis: $\mu > 40$

The statistic for this case would be:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

And replcing we got:

$t=\frac{42.6-40}{\frac{3.7}{\sqrt{15}}}=2.722$

The degrees of freedom are:

$df=n-1=15-1=14$

The p value would be:

$p_v =P(t_{(14)}>2.722)=0.0083$

We see that the p value is lower than the significance level given of 0.01 so then we can conclude that the true mean for the nicotine content is significantly higher than 40 mg

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3 years ago