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2 years ago

A function will have at most, one y-intercept. O True O False Please answer ASAP thanks

Mathematics

2 answers:

Natasha_Volkova [10]2 years ago

5 0

Answer: True. . . . . .

zmey [24]2 years ago

5 0

Answer:

It's FALSE, I guess. Cause function x=a (a=const) does not have y-intercept

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Please help, basic Algebra question

Tju [1.3M]

Answer:

The given expression $(\frac{5}{6} a^{9}p^{5}) ^{3} = \frac{125}{216} \times a^{(27) } \times p^{(15)}$

Step-by-step explanation:

Here, the given expression is: $(\frac{5}{6} a^{9}p^{5}) ^{3}$

Now, starting from the outer most bracket.

As we know :

$(abc)^{n} = (a)^{n} \times (b)^{n} \times (c)^{n}$

and $(a^m)^{n} = a^{(m \times n)}$

⇒ $(\frac{5}{6} a^{9}p^{5}) ^{3} = (\frac{5}{6})^{3} \times (a^{9})^{3} \times (p^{5}) ^{3}\\$

$=\frac{125}{216} \times (a)^{(9\times3) } \times (p)^{(5 \times 3)}\\= \frac{125}{216} \times a^{(27) } \times p^{(15)}$

Hence, the given expression $(\frac{5}{6} a^{9}p^{5}) ^{3} = \frac{125}{216} \times a^{(27) } \times p^{(15)}$

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2 years ago

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I need help I have 20 min

Bezzdna [24]

Answer:

1- d

2- a

3- d

4- b

Step-by-step explanation:

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2 years ago

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A new catalyst is being investigated for use in the production of a plastic chemical. Ten batches of the chemical are produced.

zzz [600]

Answer:

$CI=(66.54,78.46)$

Step-by-step explanation:

Given : A new catalyst is being investigated for use in the production of a plastic chemical. Ten batches of the chemical are produced. The mean yield of the 10 batches is 72.5% and the standard deviation is 5.8%. Assume the yields are independent and approximately normally distributed.

To find : A 99% confidence interval for the mean yield when the new catalyst is used ?

Solution :

Let X be the yield of the batches.

We have given, n=10 , $\bar{X}=72.5\%$ , s=5.8%

Since the size of the sample is small.

We will use the student's t statistic to construct a 995 confidence interval.

$\bar X\pm t_{n-1,\frac{\alpha}{2}}\frac{s}{\sqrt n}$

From the t-table with 9 degree of freedom for $\frac{\alpha}{2}=0.005$

$t_{n-1,\frac{\alpha}{2}}=t_{9,0.005}$

$t_{n-1,\frac{\alpha}{2}}=3.250$

The 99% confidence interval is given by,

$CI=72.5 \pm 3.25\frac{5.8}{\sqrt{10}}$

$CI=72.5 \pm 5.96$

$CI=(72.5+5.96),(72.5-5.96)$

$CI=(66.54,78.46)$

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