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3 years ago

Select the inequality below that could have 7 as a solution.

Mathematics

1 answer:

Vikentia [17]3 years ago

8 0

14x 28
There both multiples of 7

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Write the difference 16 -(-53) as a sum then simplify

valina [46]

Answer:

Step-by-step explanation:

(-)(-) = (+)

(+)(+) = (+)

(-)(+) = (+)(-) = (-)

--------------------------------------

16 - (-53) = 16 + 53 = 69

3 0

3 years ago

A data set includes 103 body temperatures of healthy adult humans having a mean of 98.1 F and a standard deviation of 0.56 F. Co

Ira Lisetskai [31]

Answer:

The 99% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.

Step-by-step explanation:

The first step is finding the confidence interval

The sample size is 103.

The first step to solve this problem is finding how many degrees of freedom there are, that is, the sample size subtracted by 1. So

$df = 103-1 = 102$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.99}{2} = \frac{0.01}{2} = 0.005$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 102 and 0.005 in the t-distribution table, we have $T = 2.63$ .

Now, we need to find the standard deviation of the sample. That is:

$s = \frac{0.56}{\sqrt{103}} = 0.055$

Now, we multiply T and s

$M = T*s = 2.63*0.055 = 0.145$

For the lower end of the interval, we subtract the mean by M. So 98.1 - 0.145 = 97.955F.

For the upper end of the interval, we add the mean to M. So 98.1 + 0.145 = 98.245F.

The 99% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.

5 0

3 years ago

Write out the first four terms of the series to show how the series starts. Then find the sum of the series or show that it dive

Nostrana [21]

Answer:

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$ = 14.25

Step-by-step explanation:

We know that

Sum of convergent series is also a convergent series.

We know that,

$\sum_{k=0}^\infty a(r)^k$

If the common ratio of a sequence |r| <1 then it is a convergent series.

The sum of the series is $\sum_{k=0}^\infty a(r)^k=\frac{a}{1-r}$

Given series,

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

$=(9+3)+(\frac97+\frac35)+(\frac9{7^2}+\frac3{5^2})+(\frac9{7^3}+\frac3{5^3})+.......$

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

Let

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$ and $t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

Now for $S_n$ ,

$S_n=9+\frac97+\frac{9}{7^2}+\frac9{7^3}+.......$

$=\sum_{n=0}^\infty9(\frac 17)^n$

It is a geometric series.

The common ratio of $S_n$ is $\frac17$

The sum of the series

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$

$=\frac{9}{1-\frac17}$

$=\frac{9}{\frac67}$

$=\frac{9\times 7}{6}$

=10.5

Now for $t_n$

$t_n= 3+\frac35+\frac{3}{5^2}+\frac3{5^3}+.......$

$=\sum_{n=0}^\infty3(\frac 15)^n$

It is a geometric series.

The common ratio of $t_n$ is $\frac15$

The sum of the series

$t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

$=\frac{3}{1-\frac15}$

$=\frac{3}{\frac45}$

$=\frac{3\times 5}{4}$

=3.75

The sum of the series is $\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

= $S_n+t_n$

=10.5+3.75

=14.25

4 0

3 years ago

A researcher would like to estimate p, the proportion of U.S. adults who support recognizing civil unions between gay or lesbian

Minchanka [31]

Answer:

Question 1:

(b) 4,445

Question 2:

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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is of:

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

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Question 1:

We have no previous estimate for the population proportion, so we use $\pi = 0.5$ .

The sample size is n for which M = 0.015. So

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

$0.015 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.015\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.015}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.015})^2$

$n = 4268$

Samples above this value should be used, and the smaller sample above this value is of 4445, so the answer is given by option b.

Question 2:

Now we find M for which $n = 2222$ .

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

$M = 1.96\sqrt{\frac{0.5*0.5}{2222}}$

$M = 0.021$

So 2.1%, and the correct answer is given by option c.

3 0

2 years ago

I need about 2 of these but pls explain so I can learn and try on my own pls

GuDViN [60]

These problems are based on the Pythagorean Theorem. The Pythagorean Theorem is a^2 + b^2 = c^2. If the two sides are equal, the triangle is a right triangle. If c^2 is less, the triangle is acute. If c^2 is more, the triangle is obtuse.

Obtuse: c^2 > a^2 + b^2

Acute: c^2 < a^2 + b^2

Right: c^2 = a^2 + b^2

#7 --- Obtuse

21^2 ___ 8^2 + 15^2

441 ___ 64 + 225

441 > 289

#8 --- Right

20^2 ___ 12^2 + 16^2

400 ___ 144 + 256

400 = 400

#9 --- Acute

6^2 ___ 4^2 + 5^2

36 ___ 16 + 25

36 < 41

Hope this helps!

8 0

3 years ago