All categories

3 years ago

What is 17/12 as a mixed number

Mathematics

2 answers:

marin [14]3 years ago

6 0

The correct answer is 1 5/12. Hope this helps.

Yuliya22 [10]3 years ago

6 0

The answer would be 1 5/12.

You might be interested in

5n= 10 2 please help with instructions

pychu [463]

5n = 10^2
n = 20 Because 5 × 20 = 100

8 0

3 years ago

If a 3-pound bunch of bananas cost $1.50, how

LUCKY_DIMON [66]

Answer:

Approximately 3 bunches would equal $4.50. So you should be able to buy at least 4 bunches which would be 10 pounds

4 0

3 years ago

Solve the following by first clearing the fractions. 1/6x+1/2=1/2x+1/2

levacccp [35]

Answer:

.166x + .5 = .5x +.5

Step-by-step explanation:

3 0

3 years ago

⦁ In a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football. Construct a 95% confiden

fomenos

Answer:

95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

Step-by-step explanation:

We are given that in a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football.

Firstly, the pivotal quantity for 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is given by;

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

where, $\hat p$ = proportion of adults in the United States whose favorite sport to watch is football in a sample of 1219 adults = $\frac{354}{1219}$

n = sample of US adults = 1291

p = population proportion of adults

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

significance level are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ]

= [ $\frac{354}{1219}-1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} }$ , $\frac{354}{1219}+1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} }$ ]

= [0.265 , 0.316]

Therefore, 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

5 0

3 years ago

For altitudes up to 36,000 feet, the relationship between ground temperature and atmospheric

Verizon [17]

Answer:

The equation for a is $a=-\frac{2000}{7}*(t-g)$

The altitute is 101,428.57 feet

Step-by-step explanation:

You know that the relationship between ground temperature and atmospheric temperature can be described by the formula

t = -0.0035a +g

where:

t is the atmospheric temperature in degrees Fahrenheit
a is the altitude, in feet, at which the atmospheric temperature is measured
g is the ground temperature in degrees Fahrenheit.

Solving the equation for a:

-0.0035a +g=t

-0.0035a= t - g

$a=\frac{t-g}{-0.0035}$

$a=-\frac{2000}{7}*(t-g)$

The equation for a is $a=-\frac{2000}{7}*(t-g)$

If the atmospheric temperature is -305 °F and the ground temperature is 50 °F, then t= -305 °F and g= 50 °F

Replacing in the equation for a you get:

$a=-\frac{2000}{7}*(-305-50)$

$a=-\frac{2000}{7}*(-355)$

a= 101,428.57

The altitute is 101,428.57 feet

4 0

3 years ago