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3 years ago

What are the values for X-ray satisfy the equation

Mathematics

1 answer:

Sonbull [250]3 years ago

5 0

Answer:

It's D

Step-by-step explanation:

you just look at the end it's says -14. u have to see what numbers should multiply in order to get negative 14, you can use -2 and 7, or -7 and 2, but you have to see what numbers from the -14 will add up -5 because u have five in there, therfore -7+2 equals -5.* This only happens if you have 1 behind the x^2.

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Someone pls help me plsss

Lelu [443]

Answer:

A) 2 5/12

Step-by-step explanation:

The word MORE refers to subtraction. If you know how to calculate fractions. Great! You should know what 2 3/4 - 1/3 equals. But if you do not:

You can use a regular calculator. Just convert everything to decimals. ( "/" means divide)

3/4 (3÷4) = 0.75 (dont forget the 2 since is has 2 3/4) so 2.75.

1/3 (1÷3) = 0.333 (underline or overline (above the number) means that number repeats. if you are using a calculator. just max the numbers out till you can put it anymore.

2.75 - 0.3 = 2.416. Now you dont know what that 416 is so just subtract the answers until it matches.

B) 1 ÷ 12 = 0.83. well it isnt 416 so that isnt right.

4 ÷ 7 = 0.571... (... means a bunch of numbers that go on and on.) again that isnt correct. Now lets try A):

5 ÷ 12 = 0.416!

Hope this helps!

~R3VO

6 0

3 years ago

Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis

lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = the interval of time between the eruption

So, X ~ Normal( $\mu=75, \sigma^{2} =20$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

P(X > 84 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{84-75}{20}$ ) = P(Z > 0.45) = 1 - P(Z $\leq$ 0.45)

= 1 - 0.6736 = 0.3264

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{13} } }$ ) = P(Z > 1.62) = 1 - P(Z $\leq$ 1.62)

= 1 - 0.9474 = 0.0526

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{20} } }$ ) = P(Z > 2.01) = 1 - P(Z $\leq$ 2.01)

= 1 - 0.9778 = 0.0222

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0

3 years ago

Of 580580 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compa

Lubov Fominskaja [6]

Answer:

a) The 99% confidence interval would be given (0.204;0.296).

b) We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).

c) No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

Step-by-step explanation:

Part a

Data given and notation

n=580 represent the random sample taken

X represent the seafood sold in the country that is mislabeled or misidentified by the people

$\hat p=0.25$ estimated proportion of seafood sold in the country that is mislabeled or misidentified by the people

$\alpha=0.01$ represent the significance level (no given, but is assumed)

p= population proportion of seafood sold in the country that is mislabeled or misidentified by the people

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".

The population proportion have the following distribution

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

The confidence interval would be given by this formula

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

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$0.25 - 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.204$

$0.25 + 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.296$

And the 99% confidence interval would be given (0.204;0.296).

Part b

We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).

Part c

A government spokesperson claimed that the sample size was too small, relative to the billions of pieces of seafood sold each year, to generalize. Is this criticism valid?

No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

4 0

4 years ago

In a relay race, the probability of the galaxy team winning is 22%. in another unrelated relay race, the probability of the kome

lina2011 [118]

We know that
the probability of the Galaxy team winning is 22%------> 0.22
the probability of the Galaxy team losing is 100%-22%----> 78%-----> 0.78

the probability of the Komets team winning is 47%------> 0.47
the probability of the Komets team losing is 100%-47%----> 53%-----> 0.53

therefore
the probability of the comets losing their race and the galaxy winning theirs is
0.53*0.22=0.1166------> 11.66%

the answer is
11.66%

4 0

3 years ago

Fifteen less than eight times a number (n) is twenty seven in equation form

Zepler [3.9K]

Answer:

x = -4/5 = -0.800

Step-by-step explanation:

Pull out like factors :

-12 - 15x = -3 • (5x + 4)

Solve :

-3 = 0

Solve :

5x+4 = 0

5x = -4

x = -4/5 = -0.800

5 0

4 years ago