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2 years ago

7, 16, 25, 34, 43, ..... nth term is

Mathematics

2 answers:

serious [3.7K]2 years ago

7 0

$\bf 7~~,~~\stackrel{7+9}{16}~~,~~\stackrel{16+9}{25}~~,~~\stackrel{25+9}{34}~~,~~\stackrel{34+9}{43}$

so, as we can see, we're really adding 9 to the current term, to get the next term, therefore, 9 is the common difference, and of course 7 is the first term.

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\a_n=a_1+(n-1)d\qquad\begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\d=\textit{common difference}\\[-0.5em]\hrulefill\\d=9\\a_1=7\end{cases}\\\\\\a_n=7+(n-1)9$

vladimir2022 [97]2 years ago

6 0

Answer:

79 is the ninth term.

Step-by-step explanation:

You add nine to each one.

7, 16, 25, 34, 43, 52, 61, 70, 79

79 is the ninth term.

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Answer:

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Step-by-step explanation:

10 tbsp of flour

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2 years ago

PLEASE HELP, I’ve been stuck on this question for a whileIn the diagram, AB is divided into equal parts. The coordinates of poin

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AB is divided into 8 equal parts and point C is 1 part FROM A TO B, so the ratio is 1:7, with C being 1/7 of the way. The ratio is k, found by writing the numerator of the ratio (1) over the sum of the numerator and denominator (1+7). So our k value is 1/8. Now we need to find the rise and the run (slope) of the points A and B. $m= \frac{5-9}{9-(-3)}$ . That gives us a rise of -4 and a run of 12. The coordinates of C are found in this formula: $C(x,y)=[ x_{1} +k(run), y_{1} +k(rise)]$ . Filling in accordingly, we have $C(x,y)=[-3+ \frac{1}{8}(12),9+ \frac{1}{8}(-4)]$ which simplifies a bit to $C(x,y)=(-3+ \frac{3}{2},9- \frac{1}{2})$ . Finding common denominators and doing the math gives us that the coordinates of point C are $(- \frac{3}{2}, \frac{17}{2})$ . There you go!

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Which measure is the largest? 1.2 km, 120,600 cm, 1,220,000 mm, 120 meters

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The answer would be 1,220,000mm.

You can do this if you convert all the measures into one unit. Let us convert all into km.

The first one is already in km, so we do not need to covert it.

Let's start with converting 1,220,000mm to km.

There are 1, 000,000 mm in one km.

$1,220,000X \frac{1km}{1,000,000mm} = 1.22km$

So there are 1.22km in 1,220,000mm.

Next, we have 120m. There are 1,000m in 1km.

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There are 0.12km in 120m

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2 years ago

A large manufacturing plant has analyzed the amount of time required to produce an electrical part and determined that the times

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Answer:

We conclude that the new procedure will not decrease the population mean amount of time required to produce the part.

Step-by-step explanation:

We are given that a large manufacturing plant has analyzed the amount of time required to produce an electrical part and determined that the times follow a normal distribution with mean time μ = 45 hours.

A random sample of 25 parts will be selected and the average amount of time required to produce them will be determined. The sample mean amount of time is = 43.118 hours with the sample standard deviation s = 5.5 hours

Let $\mu$ = population mean amount of time required to produce an electrical part using new procedure

SO, Null Hypothesis, $H_0$ : $\mu \geq$ 45 hours {means that the new procedure will remain same or increase the population mean amount of time required to produce the part}

Alternate Hypothesis, $H_a$ : $\mu$ < 45 hours {means that the new procedure will decrease the population mean amount of time required to produce the part}

The test statistics that will be used here is One-sample t test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{{\frac{s}{\sqrt{n} } } }$ ~ $t_n_-_1$

where, $\mu$ = sample mean amount of time = 43.118 hours

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So, test statistics = $\frac{43.118-45}{{\frac{5.5}{\sqrt{25} } } }$ ~ $t_2_4$

= -1.711

Now at 0.025 significance level, the t table gives critical value of -2.064 at 24 degree of freedom for left-tailed test. Since our test statistics is more than the critical value of t so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the new procedure will remain same or increase the population mean amount of time required to produce the part.

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