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Free_Kalibri [48]

2 years ago

14

In Texas, the coldest temperature ever recorded is -23 F. That's a lot chillier than in Florida, where the coldest recorded temp

erature is -2 F. How many degrees warmer is Florida's coldest recorded temperature than Texas's?

2 answers:

Semmy [17]2 years ago

6 0

Answer:

21

Step-by-step explanation:

The answer is negitive 21 becuase it you sub and

beks73 [17]2 years ago

3 0

Answer:

21

Step-by-step explanation:

21+-23=-2

You might be interested in

Zöe schedules advertising for a radio station. She must fill 12 minutes each hour with 30 second ads and 60 second ads. Zöe sold

PilotLPTM [1.2K]

I cannot see Zoe's work to explain the error, but the correct method of solving is listed:

x is the number of 30-second ads
y is the number of 60-second ads

x+y=12(60)=720 would be the first equation; this is because while the ads together make 12 minutes, the ad times are in seconds. This means we must multiply 12 by 60.

y=2x is the second equation

Our system is then
x+y=720
y=2x

We will use substitution to solve this. Plug 2x in place of y in the first equation:
x+2x = 720

Combine like terms:
3x = 720

Divide both sides by 3:
3x/3 = 720/3
x = 240

Substitute this value in for x in the second equation:
y=2(240)
y=480

8 0

2 years ago

A brewery produces cans of beer that are supposed to contain exactly 12 ounces. But owing to the inevitable variation in the fil

MArishka [77]

Answer:

$T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)$

$P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z$

Step-by-step explanation:

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solutio to the problem

Let X the random variable that represent the amount of beer in each can of a population, and for this case we know the distribution for X is given by:

$X \sim N(12,0.3)$

Where $\mu=12$ and $\sigma=0.3$

For this case we select 6 cans and we are interested in the probability that the total would be less or equal than 72 ounces. So we need to find a distribution for the total.

The definition of sample mean is given by:

$\bar X = \frac{\sum_{i=1}^n X_i}{n} = \frac{T}{n}$

If we solve for the total T we got:

$T= n \bar X$

For this case then the expected value and variance are given by:

$E(T) = n E(\bar X) =n \mu$

$Var(T) = n^2 Var(\bar X)= n^2 \frac{\sigma^2}{n}= n \sigma^2$

And the deviation is just:

$Sd(T) = \sqrt{n} \sigma$

So then the distribution for the total would be also normal and given by:

$T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)$

And we want this probability:

$P(T\leq 72)$

And we can use the z score formula given by:

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

$P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z$

6 0

2 years ago

A pack of 6 batteries of the same type costs £2.79. By caculating the price, per battery, determine if this is better value. You

Fittoniya [83]

Answer:

£0.465. It is a better value.

Step-by-step explanation:

If a pack of 6 batteries costs £2.79, price of a battery can be calculated as shown;

6batteries = £2.79

1 battery = x

x = cost of a battery

6x = £2.79

x = £2.79/6

x = £0.465

Price per battery will be £0.465. This price is a better value since the price is lesser than the cost of the 6batteries (£2.79) in a pack.

6 0

2 years ago

Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>

<em />

<u>So, 90% confidence interval for the population proportion, p is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

<u>90% confidence interval for p</u> = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

$0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }$

$\sqrt{n} = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}$

$\sqrt{n}$ = 54.79

n = $54.79^{2}$

n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

5 0

2 years ago

WILL MARK BRANLIESTHELP ME PLZ

laiz [17]

I believe it’s 0.043

6 0

2 years ago

Read 2 more answers