Need help with this questions please.

If the maximum value of y=-3x² + mx + 10 is 37, find the values of m.

uysha [10]

Step-by-step explanation:

y = -3x² + mx + 10

dy/dx = -6x + m.

When dy/dx = 0, -6x + m = 0. => x = m/6.

We have y = -3(m/6)² + m(m/6) + 10 = 37.

=> -m²/12 + m²/6 = 27

=> -m² + 2m² = 324

=> m² = 324

=> m = 18 or m = -18.

5 0

2 years ago

Read 2 more answers

I need with this question please. Please help me

klemol [59]

Answer:

idk

Step-by-step explanation:

to lazy

7 0

3 years ago

Write 5 pounds for $2.49 as a unit rate. round to the nearest hundreth

DaniilM [7]

5 pounds = $2.49

1 pound = $2.49 ÷ 5 = $0.50 (nearest hundredth)

.

Answer: $0.50/pound

5 0

2 years ago

Read 2 more answers

A self proclaimed psychic was tested for ESP. The psychic was presented with 200 cards face down and asked to determine if the c

swat32

Answer:

The 95% confidence interval would be given (0.172;0.288).

We are confident at 95% that the true probability that the psychic correctly identifies the symbol on the card in a random trial is between (0.172;0.288).

We can conclude that her random guessing would have got her correct 20% of the time. Since the confidence interval contains 0.2, she has no psychic powers, rather she just guessed it like normal people. Any person could have got it correct this many times.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Description in words of the parameter p

$p$ represent the probability that the psychic correctly identifies the symbol on the card in a random trial

$\hat p$ represent the estimated probability that the psychic correctly identifies the symbol on the card in a random trial

n=200 is the sample size required

$z_{\alpha/2}$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Numerical estimate for p

In order to estimate a proportion we use this formula:

$\hat p =\frac{X}{n}$ where X represent the cases successful

$\hat p=\frac{46}{200}=0.23$ represent the estimated probability that the psychic correctly identifies the symbol on the card in a random trial

Confidence interval

The confidence interval for a proportion is given by this formula:

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.23 - 1.96 \sqrt{\frac{0.23(1-0.23)}{200}}=0.172$

$0.23 + 1.96 \sqrt{\frac{0.23(1-0.23)}{200}}=0.288$

And the 95% confidence interval would be given (0.172;0.288).

We are confident at 95% that the true probability that the psychic correctly identifies the symbol on the card in a random trial is between (0.172;0.288).

We can conclude that her random guessing would have got her correct 20% of the time. Since the confidence interval contains 0.2, she has no psychic powers, rather she just guessed it like normal people. Any person could have got it correct this many times.

6 0

3 years ago

Past records indicate that the probability of online retail orders that turn out to be fraudulent is 0.08. Suppose that, on a gi

Sunny_sXe [5.5K]

Answer:

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

Step-by-step explanation:

We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.08.

The probability of k online retail orders that turn out to be fraudulent in the sample is:

$P(x=k)=\dbinom{n}{k}p^k(1-p)^{n-k}=\dbinom{20}{k}\cdot0.08^k\cdot0.92^{20-k}$

We have to calculate the probability that 2 or more online retail orders that turn out to be fraudulent. This can be calculated as:

$P(x\geq2)=1-[P(x=0)+P(x=1)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.08^{0}\cdot0.92^{20}=1\cdot1\cdot0.189=0.189\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.08^{1}\cdot0.92^{19}=20\cdot0.08\cdot0.205=0.328\\\\\\\\P(x\geq2)=1-[0.189+0.328]\\\\P(x\geq2)=1-0.517=0.483$

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

5 0

3 years ago