Seven days ago, I put 60 jelly beans in a bowl. Now there are 4 jelly beans left.

Exponent rules . Simplify each expression . HELPPP pls on these 2 problems

Vedmedyk [2.9K]

Step-by-step explanation:

Please refer to the attachment

8 0

2 years ago

Help me with this pls

Ne4ueva [31]

Answer:

The answer should be C) 4x + 5x = 180

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

The area of a rectangular wall of a barn is 140 square feet. Its length is 6 feet longer than twice its width. Find the length a

Lubov Fominskaja [6]

Do you need to graph it?

3 0

2 years ago

In a box, 1/4 of the beads are red, 3/5 of the remainder are yellow, and the rest are blue. If there are 48 blue beads, how many

Andrei [34K]

Answer:

160 total beads

Step-by-step explanation:

so if 1/4 of the beads are red, then 3/4 of them are not.....so 3/4 is the remainder of beads.....and 3/5 of the remainder are yellow....so 3/5 of 3/4 =

3/5 * 3/4 = 9/20...so 9/20 are yellow.....and the rest (48) are blue.

1/4 + 9/20 = 5/20 + 9/20 = 14/20 reduces to 7/10...so 7/10 of the beads are red and yellow

so if 7/10 of the beads are red and yellow, then 3/10 are blue

3/10 of what number is 48

3/10x = 48

x = 48 * 10/3

x = 480/3

x = 160

let me check it..

1/4 are red......160 total beads.....so red beads = (1/4 * 160) = 160/4 = 40

3/5 of the remainder is yellow.....so 3/5 of (160 - 40) = 3/5(120) = 72 yellow

and then u have 48 blue...

40 + 72 + 48 = 160

so there are 160 total beads.......40 red, 72 yellow, and 48 blue <===

6 0

3 years ago

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x overbar, is found

joja [24]

Answer:

(a) 80% confidence interval for the population mean is [109.24 , 116.76].

(b) 80% confidence interval for the population mean is [109.86 , 116.14].

(c) 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed.

Step-by-step explanation:

We are given that a simple random sample of size n is drawn from a population that is normally distributed.

The sample mean is found to be 113 and the sample standard deviation is found to be 10.

(a) The sample size given is n = 13.

Firstly, the pivotal quantity for 80% 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 mean = 113

s = sample standard deviation = 10

n = sample size = 13

$\mu$ = population mean

Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 80% confidence interval for the population mean, $\mu$ is ;

P(-1.356 < $t_1_2$ < 1.356) = 0.80 {As the critical value of t at 12 degree

of freedom are -1.356 & 1.356 with P = 10%}

P(-1.356 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.356) = 0.80

P( $-1.356 \times }{\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.356 \times }{\frac{10}{\sqrt{13} } }$ , $113+1.356 \times }{\frac{10}{\sqrt{13} } }$ ]

= [109.24 , 116.76]

Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].

(b) Now, the sample size has been changed to 18, i.e; n = 18.

So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.

80% confidence interval for $\mu$ = [ $\bar X-1.333 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.333 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.333 \times }{\frac{10}{\sqrt{18} } }$ , $113+1.333 \times }{\frac{10}{\sqrt{18} } }$ ]

= [109.86 , 116.14]

Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].

(c) Now, we have to construct 98% confidence interval with sample size, n = 13.

So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.

98% confidence interval for $\mu$ = [ $\bar X-2.681 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+2.681 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-2.681 \times }{\frac{10}{\sqrt{13} } }$ , $113+2.681 \times }{\frac{10}{\sqrt{13} } }$ ]

= [105.56 , 120.44]

Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed because t test statistics is used only when the data follows normal distribution.

6 0

2 years ago