3 years ago

Simplify the expression 14b-2-17b

Mathematics

1 answer:

Kamila [148]3 years ago

4 0

The answer to the following question that you have asked is -2-3b

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A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 48 specimens and counts the number of

baherus [9]

Answer: The open interval would be (31.4,42.5).

Step-by-step explanation:

Since we have given that

mean = 36.9

Standard deviation = 16.5

n = 48

At 98% confidence interval, z = 2.33

So, Interval would be

$\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=36.9\pm 2.33\dfrac{16.5}{\sqrt{48}}\\\\=36.9\pm 5.549\\\\=(36.9-5.5,36.9+5.6\\\\=(31.4,42.5)$

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3 years ago

Assuming that the sample mean carapace length is greater than 3.39 inches, what is the probability that the sample mean carapace

joja [24]

Answer:

The answer is "".

Step-by-step explanation:

Please find the complete question in the attached file.

We select a sample size n from the confidence interval with the mean $\mu$ and default $\sigma$ , then the mean take seriously given as the straight line with a z score given by the confidence interval

$\mu=3.87\\\\\sigma=2.01\\\\n=110\\\\$

Using formula:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

The probability that perhaps the mean shells length of the sample is over 4.03 pounds is

$P(X>4.03)=P(z>\frac{4.03-3.87}{\frac{2.01}{\sqrt{110}}})=P(z>0.8349)$

Now, we utilize z to get the likelihood, and we use the Excel function for a more exact distribution

$=\textup{NORM.S.DIST(0.8349,TRUE)}\\\\P(z$

the required probability: $P(z>0.8349)=1-P(z$

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2 years ago

Jacob has $38 to spend on a field trip to the state museum. He spends $12 on admission to the museum. In the souvenir store he f

netineya [11]

Answer:

the cap and the pic EACH cost $13

Step-by-step explanation:

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3 years ago