573 dividing 3 deciding were to start dividing

Rent and other associated housing costs, such as utilities, are an important part of the estimated costs of attendance at colleg

zhenek [66]

Answer:

96% confidence interval estimate for the mean monthly rent of all unmarried BYU students in winter 2018 is [335.89 , 356.10].

Step-by-step explanation:

We are given that a group of researchers at the BYU Off-Campus Housing department want to estimate the mean monthly rent that unmarried BYU students paid during winter 2018.

During March 2018, they randomly sampled 314 BYU students and found that on average, students paid $346 for rent with a standard deviation of $86.

So, the pivotal quantity for 95% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = average rent paid by 314 BYU students = $346

s = sample standard deviation = $86

n = sample of students = 314

$\mu$ = population mean monthly rent of all unmarried BYU students

So, 96% confidence interval for the population mean monthly rent, $\mu$ is ;

P(-2.082 < $t_3_1_3$ < 2.082) = 0.96

P(-2.082 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.082) = 0.96

P( $-2.082 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.082 \times {\frac{s}{\sqrt{n} } }$ ) = 0.96

P( $\bar X -2.082 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X +2.082 \times {\frac{s}{\sqrt{n} } }$ ) = 0.96

96% confidence interval for $\mu$ = [ $\bar X -2.082 \times {\frac{s}{\sqrt{n} } }$ , $\bar X +2.082 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $346 -2.082 \times {\frac{86}{\sqrt{314} } }$ , $346 +2.082 \times {\frac{86}{\sqrt{314} } }$ ]

= [335.89 , 356.10]

Therefore, 96% confidence interval estimate for the mean monthly rent of all unmarried BYU students in winter 2018 is [335.89 , 356.10].

5 0

3 years ago

Help pls ty in advance

svetoff [14.1K]

$\bold{\huge{\underline{ Solution\: 2 }}}$

The name of the circle is IJKL
The name of central Angle
$\sf{\underline{= {\angle} IHJ }}$
The name of the secant = KL

${\bigstar}$ Circle :- It is a round figures, which has no end points.

${\bigstar}$ Central angle :- It the angle ,which lie in the middle of the circle.

Here, H is the mid point of the circle .So, Angle IHJ is the central angle.

${\bigstar}$ Secant :- A line drawn in circle in such a way that it intersect the circle from two distinct points. Then, It is called secant .

Here, KL is the line which acts as a secant because it intersect the circle from point K and point L.

$\bold{\huge{\underline{ Solution\: 3 }}}$

Semicircle = FGH and FIH
Minor Arc :- FG or GH
Major Arc :- FHG or GFH

${\bigstar}$ Semicircle :- Semicircle is nothing but the half of the circle.

Here, In second diagram 

FGH and FIH are acting as two semicircles which simulatineously forming one circle.

${\bigstar}$ Minor Arc :- Minor Arc is nothing but the smaller arc of the circle and it is always smaller than the half of the circle that is semicircle.

Here, 

FG or GH are the minor arc of the given circle.

${\bigstar}$ Major Arc :- Major Arc is nothing but the largest arc of the circle and it is always larger than the half of the circle that is semicircle.

Here, 

FHG or GFH is the major arc as it is larger than the semicircle of the given circle .

7 0

2 years ago

Una heladeria dispone de 20 frutas distintas para elaborar sus malteadas. Si los clientes pueden elegir tres sabores para mezcla

Dahasolnce [82]

Answer:

Existen 6840 permitaciones de malteadas de tres sabores distintos que la heladería puede ofrecer.

Step-by-step explanation:

En este caso, el cliente que adquiere una malteada de tres sabores distintos sigue el siguiente procedimiento:

1) El primer sabor sale de cualquiera de las 20 frutas disponibles.

2) El segundo sabor es distinto al primer sabor, es decir, que sale de las 19 frutas restantes.

3) El tercer sabor es distinto al primer sabor y al segundo sabor, es decir, que sale de las 18 frutas restantes.

Puesto que existe una doble conjunción y que puede importar el orden según la preferencia del cliente, se habla matemáticamente de una permutación, definida como:

$n\mathbb{P}k = \frac{n!}{(n-k)!}$ (1)