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mr Goodwill [35]

3 years ago

14

A nutrition lab tested 40 hot dogs to see if their mean sodium content was less than the 325-mg upper limit set by regulations f

or "reduced sodium" franks. A 90% confidence interval estimated the mean sodium content for this kind of hot dog at 317.2 to 326.8 mg. Given this, what would you tell the lab about whether the hot dogs satisfy the regulation.

1 answer:

tekilochka [14]3 years ago

4 0

Answer:

Step-by-step explanation:

Given that a nutrition lab tested 40 hot dogs to see if their mean sodium content was less than the 325-mg upper limit set by regulations for "reduced sodium" franks

90% confidence interval is

(317.2, 326.8)mg.

Using confidence intervals,we can check whether null hypothesis is true.

Here

$H_0: \bar x = 325\\H_a: \bar x$

(left tailed test at 10% significance level)

We check whether confidence interval 90% contains the mean value.

We find 325 belongs to the confidence interval

Hence H0 is true

The hot dogs do not satisfy.

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Eva8 [605]

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4 0

3 years ago

Read 2 more answers

What is the nth term?

kumpel [21]

Let $a_k$ denote the <em>k</em>th term of the sequence. Then

$a_k=a_1+d(k-1)$

where <em>d</em> is the common difference between consecutive terms in the sequence and <em>a</em>₁ is the first term.

The sum of the first <em>n</em> terms is

$S_n=\displaystyle\sum_{k=1}^na_k=a_1+a_2+\cdots+a_{n-1}+a_n$

From the formula for $a_k$ , we get

$S_n=\displaystyle\sum_{k=1}^n(a_1+d(k-1))=a_1\sum_{k=1}^n1+d\sum_{k=1}^n(k-1)$

$S_n=\displaystyle na_1+d\sum_{k=0}^{n-1}k$

$S_n=na_1+\dfrac{d(n-1)n}2$

$S_n=\dfrac n2(2a_1-d+dn)$

So we have $d=-5$ , and $2a_1-d=16$ so that $a_1=\frac{11}2$ .

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$a_n=\dfrac{11}2-5(n-1)=\boxed{\dfrac{21-10n}2}$

7 0

3 years ago

Janaya buys a postcard from each travel destination she visits. Before starting her travels this year, Janaya had 20 postcards.

MrRa [10]

Answer:

The percentage increase in Jayana's postcard collection is 15%

Step-by-step explanation:

Jayana had 20 postcards last year

With 3 more postcards this year

From the given equation Quantity = Percent x Whole

Here;

Let P = Percent/100%

Quantity = Increment in postcard = 3 postcards

Whole = The total postcards from last year = 20 postcards

Hence,

3 postcards = P x 20 postcards

P = 3/20 = 0.15

Recall P = Percent/100%

∴ Percent = P x 100% = 0.15 x 100% = 15%

The percentage increase in Jayana's postcard collection is 15%

4 0

2 years ago

Which equation represents a circle with a center at (-3,-5) and a radius of 6 units

TEA [102]

Answer:

-3x^2 - 5y^2 = 36

Step-by-step explanation:

x^2+y^2=r^2

7 0

3 years ago

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$

From equation(1), substituting x with (4x), Then:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}$

The first three terms of cos (4x) is:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}$

$\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}$

$\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}$

Multiplying equation (2) with (3); we have :

$\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Finally , multiplying 5 with $\mathtt{ e^{-x^2} cos (4x) }$ ; we have:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

7 0

3 years ago