All categories

3 years ago

You are going to the grocery to buy the candies for you and your sisters/brothers. You want to buy

Mathematics

1 answer:

Llana [10]3 years ago

5 0

Answer: 35 chocolates were bought and 52 gums were bought.

Step-by-step explanation:

c + g = 87 where c is the number of chocolate and g is the number og gums

1.50c + 0.50g = 78.50 solve both equations by elimination.

-1.50c - 1.50g = -130.5

1.50 + 0.50g = 78.50

-1g = -52

g= 52

1.50c + 0.50(52) = 78.50

1.50c + 26 = 78.50

-26 -26

1.50c = 52.5

c = 35

You might be interested in

Cost of a pen o.95 mark up 60

Alika [10]

I think the answer is 5700

8 0

3 years ago

Construct a 99% confidence interval for the population mean, mu. Assume the population has a normal distribution. A group of 1

Zarrin [17]

Answer:

99% confidence interval for the population mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% 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 age of selected students = 22.4 years

s = sample standard deviation = 3.8 years

n = sample of students = 19

$\mu$ = population mean

<em>Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.878 < $t_1_8$ < 2.878) = 0.99 {As the critical value of t at 18 degree of

freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.878) = 0.99

P( $-2.878 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.878 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ) = 0.99

<u>99% confidence interval for</u> $\mu$ = [ $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ , $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ]

= [ $22.4 -2.878 \times {\frac{3.8}{\sqrt{19} }$ , $22.4 +2.878 \times {\frac{3.8}{\sqrt{19} }$ ]

= [19.891 , 24.909]

Therefore, 99% confidence interval for the population mean is [19.891 , 24.909].

6 0

3 years ago

6. – 4y + 8x = 32 for y

MatroZZZ [7]

Answer:

The answer is: y=2x−8

Step-by-step explanation:

Step 1: Add -8x to both sides.

8x−4y+−8x=32+−8x

−4y=−8x+32

Step 2: Divide both sides by -4.

−4y−4=−8x+32−4

y=2x−8

4 0

3 years ago

Read 2 more answers

Help please...no links...

Likurg_2 [28]

the answer is 3/4 or c.

Step-by-step explanation:

to find tan A you would use the equation of tan which is the value opposite the angle A which is 27 over the value adjacent to the angle which is 36.

3 0

3 years ago

Assume that a sample is used to estimate a population proportion p. find the margin of error e that corresponds to the given sta

Leno4ka [110]

Let p be the proportion. Let c be the given confidence level , n be the sample size.

Given: p=0.3, n=1180, c=0.99

The formula to find the Margin of error is

ME = $z _{\alpha/2} \sqrt{\frac{p*(1-p)}{n}}$

Where z (α/2) is critical value of z.

P(Z < z) = α/2

where α/2 = (1- 0.99) /2 = 0.005

P(Z < z) = 0.005

So in z score table look for probability exactly or close to 0.005 . There is no exact 0.005 probability value in z score table. However there two close values 0.0051 and 0.0049 . It means our required 0.005 value lies between these two probability values.

The z score corresponding to 0.0051 is -2.57 and 0.0049 is -2.58. So the required z score will be average of -2.57 and -2.58

(-2.57) + (-2.58) = -5.15

-5.15/2 = -2.575

For computing margin of error consider positive z score value which is 2.575

The margin of error will be

ME = $z _{\alpha/2} \sqrt{\frac{p*(1-p)}{n}}$

= $2.575 \sqrt{\frac{0.30*(1-0.3)}{1180}}$

= 2.575 * 0.0133

ME = 0.0342

The margin of error is 0.0342

7 0

3 years ago