2. Consider a function f(x) that takes on values | x +2| -5 for x greater than or

A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals

andrew11 [14]

Answer:

A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle? Here the number of animals and the number of days are in inverse proportion. Hence the food will last 4 days.

8 0

3 years ago

Morgan rewrote the expression 90 + (10 + 4.8) as (90 + 10) + 4.8. How will this change the work that is needed to evaluate the e

serg [7]

Answer:

Answer is c

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

B. Rob would like to create a triangle that is

Phantasy [73]

Answer:

$14y + 9$

Step-by-step explanation:

Perimeter of the given triangular above = sum of all its sides.

$= (y + 9) + 2y + (4y - 1) = y + 9 + 2y + 4y -1$

Collect like terms

$= y + 2y + 4y + 9 - 1$

$= 7y + 8$

If Bob creates a new triangle that is 7 less than twice of the perimeter of the triangle above, the expression of the new triangle would be:

$2(7y + 8) - 7$

Simplify

$2*7y + 2*8 - 7$

$14y + 16 - 7$

New perimeter = $14y + 9$

5 0

3 years ago

An automobile company wants to determine the average amount of time it takes a machine to assemble a car. A sample of 40 times y

aksik [14]

Answer:

A 98% confidence interval for the mean assembly time is [21.34, 26.49] .

Step-by-step explanation:

We are given that a sample of 40 times yielded an average time of 23.92 minutes, with a sample standard deviation of 6.72 minutes.

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

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

where, $\bar X$ = sample average time = 23.92 minutes

s = sample standard deviation = 6.72 minutes

n = sample of times = 40

$\mu$ = population mean assembly time

Here for constructing a 98% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, a 98% confidence interval for the population mean, $\mu$ is;

P(-2.426 < $t_3_9$ < 2.426) = 0.98 {As the critical value of z at 1% level

of significance are -2.426 & 2.426}

P(-2.426 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.426) = 0.98

P( $-2.426 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.426 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

P( $\bar X-2.426 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.426 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.426 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.426 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $23.92-2.426 \times {\frac{6.72}{\sqrt{40} } }$ , $23.92+2.426 \times {\frac{6.72}{\sqrt{40} } }$ ]

= [21.34, 26.49]

Therefore, a 98% confidence interval for the mean assembly time is [21.34, 26.49] .

7 0

3 years ago

How many 4 digit numbers are there with the sum of the four digits equal to 34

Masteriza [31]

Answer:

Step-by-step explanation:

THE LARGEST 4 DIGIT NUMBER IS 9999 WHICH HAS SUM OF DIGITS 9+9+9+9 =36 SO TO HAVE A SUM OF DIGITS 34 ,WE HAVE TO TAKE 2 AWAY THERE ARE TWO WAYS TO DO THAT IS

1) TAKE 2 AWAY FROM 1 OF THE 9"S AND THE SET OF THE DIGITS (7,9,9,9)

2)TAKE 1 AWAY FROM 2 OF THE 9"S AND THE SET OF DIGITS (8,8,9,9)

THERE ARE 4 ARRANGEMENTS OF THE DIGITS (7,9,9,9) BECAUSE THERE ARE FOUR POSITIONS FOR THE DIGITS THOUSANDS,HUNDREDES,TENS AND UNITS SO WE CAN PICK A P[OSITION FOR THE 7 IN (4,1) = FOUR WAYS ( THERE ARE 7,9,9,9, 9,7,9,9, 9,9,7,9, 9,9,9,7

THERE ARE SIX ARRANGEMENTS OF THE DIGITS (8,8,9,9, BECAUSE THERE ARE FOUR POSITIONF FOR THE DIGITS THOUSANDS , HUNDRED , TENS AND UNITS SO WE CAN PICK A POSITIONS FOR THE 28"S IN (4,2) = 6 WAYS THEY ARE 8,8,9,9, 8,9,8,9, 8,9,9,8, SO THAT 4+6 OR 10 WAYS

3 0

3 years ago