Select all x values that are solutions to the equation

Jack picked 76 apples. Of these, he sold 4y apples to Juanita and 3t apples to Sylvia. If he kept the remaining apples, how many

Alex Ar [27]

Answer:

Number of apples remaining with him will be 2 when we combined both the conditions.

Step-by-step explanation:

Given;

Number of apples Jack Picked =76

Number of apples sold to Juanita = 4y

Number of apples sold to Sylvia = 3t

We need to find the Number of apples remaining with him.

Solution:

To find the Number of apples remaining with him we will Subtract Number of apples sold to Juanita and Number of apples sold to Sylvia from Number of apples Jack Picked.

framing in equation form we get;

Number of apples remaining with him = $76-4y-3t$

Now Given:

Condition 1 : y ≥ 15 and t = 2

So we can assume y =16 and t =2

Number of apples remaining with him = $76-4\times16-3\times2 = 76 -64 -6 = 6$

Also we can say that y =17 and t =2

Number of apples remaining with him = $76-4\times17-3\times2 = 76 -68 -6 = 2$

So the Given data is Insufficient to find the Number of apples remaining with him.

Condition 2 : y = 17

Number of apples remaining with him = $76-4\times17-3t = 76 -68 -3t = 8-3t$

Since value of t is not given we can say that;

Given data is insufficient to find the number of apples remaining with him.

Condition 3 : Combined i.e y = 17 and y≥ 15 t = 2

Number of apples remaining with him = $76-4\times17-3\times2 = 76 -68 -6 = 2$

Hence we can say that Number of apples remaining with him will be 2 when we combined both the conditions.

7 0

3 years ago

Suppose a simple random sample of size nequals81 is obtained from a population with mu equals 79 and sigma equals 18. (a) Descr

Daniel [21]

Answer:

(a) The sampling distribution of $\overline{X}$ = Population mean = 79

(b) P ( $\overline{X}$ greater than 81.2 ) = 0.1357

(c) P ( $\overline{X}$ less than or equals 74.4 ) = .0107

(d) P (77.6 less than $\overline{X}$ less than 83.2 ) = .7401

Step-by-step explanation:

Given -

Sample size ( n ) = 81

Population mean $(\nu)$ = 79

Standard deviation $(\sigma )$ = 18

(a) Describe the sampling distribution of $\overline{X}$

For large sample using central limit theorem

the sampling distribution of $\overline{X}$ = Population mean = 79

(b) What is Upper P ( $\overline{X}$ greater than 81.2 ) =

$P(\overline{X}> 81.2)$ = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}> \frac{81.2 - 79}{\frac{18}{\sqrt{81}}})$

= $P(Z> 1.1)$

= $1 - P(Z< 1.1)$

= 1 - .8643 =

= 0.1357

(c) What is Upper P ( $\overline{X}$ less than or equals 74.4 ) =

$P(\overline{X}\leq 74.4)$ = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq \frac{74.4- 79}{\frac{18}{\sqrt{81}}})$

= $P(Z\leq -2.3)$

= .0107

(d) What is Upper P (77.6 less than $\overline{X}$ less than 83.2 ) =

$P(77.6< \overline{X}< 83.2)$ = $P(\frac{77.6- 79}{\frac{18}{\sqrt{81}}})< P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq \frac{83.2- 79}{\frac{18}{\sqrt{81}}})$