What two integers does the square root of 429 lie?

Door panels are stored in the yard on pallets holding 20 panels. The yard currently has 5 complete pallets and 20 additional pal

Sladkaya [172]

At the end of tomorrow, inventory of 5+20 pallets will have been used to fill orders for 2+10 pallets. That leaves inventory on hand of

... 25 - 12 = 13 pallets

Since each holds 20 panels, that amounts to

... (13 pallets) × (20 panels/pallet) = 260 panels . . . . remaining

5 0

2 years ago

Jane invested £4000 for 3 years with an interest rate of 1.5%. What was her investment worth at the end of this period?

ddd [48]

Answer:

$4182.7

Step-by-step explanation:= 4000*(1.5%)*3

Year 1= 4000*(100%+1.5%)= 4060

Year 2= 4060*(100%+1.5%)= 4120.9

Year 3= 4120.9*(100%+1.5%)= 4182.7

6 0

2 years ago

Read 2 more answers

Suppose that the weight of seedless watermelons is normally distributed with mean 6.1 kg. and standard deviation 1.9 kg. Let X b

horrorfan [7]

Answer:

a. $X\sim N(\mu = 6.1, \sigma = 1.9)$ b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349

Step-by-step explanation:

a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.

b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.

c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842

d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166

e. The z-score related to 6.4 kg is $z_{1} = (6.4-6.1)/1.9 = 0.1579$ and the z-score related to 7 kg is $z_{2} = (7-6.1)/1.9 = 0.4737$ , we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194

f. The 90th percentile for the standard normal distribution is 1.2815, therefore, the 90th percentile for the given distribution is 6.1 + (1.2815)(1.9) = 8.5349

7 0

2 years ago

Exhibit 11-2 We are interested in determining whether the variances of the sales at two music stores (A and B) are equal. A samp

Fiesta28 [93]

Answer:

Option A , should not be rejected

Step-by-step explanation:

From the question we are told that:

Sample 1 $n=25$

Standard deviation 1 $\sigma_1=30$

Sample 2 $n=16$

Standard deviation 2 $\sigma_2=20$

Level of confidence $\alpha=95\%=>0.95$

Generally The Hypothesis are given as

$H_0:\sigma_1^2=\sigma_1^2$

$H_1:\sigma_1^2\neq \sigma_1^2$

Generally the equation for test Statistics is mathematically given by

$T=\frac{\sigma_1^2}{\sigma_2^2}$

$T=\frac{30^2}{20^2}$

$T=2.25$

Therefore

Critical Value

$X=(\alpha,df_1)$

$X'=(\alpha,df_2)$

Where

$df=n-1$

Therefore

$X=(0.95,24)$

$X'=(0.95,15)$

From Table

$T_{Critical}=T_x,T_{x'}$

$T_{Critical}=0.41,2.701$

Therefore

$T_x$

Hence,We fail to reject the Null Hypothesis $H_0$

Option A , should not be rejected

3 0

1 year ago

1. Contaminated water is subjected to a cleaning process. The concentration of the pollutants is initially 5 mg per liter of wat

Art [367]

Answer:

$C(t)=5\cdot(0.9)^t$

Step-by-step explanation:

The exponential function is often used to model natural growing or decaying processes, where the change is proportional to the actual quantity.

An exponential decaying function is expressed as:

$C(t)=C_o\cdot(1-r)^t$

Where:

C(t) is the actual value of the function at time t

Co is the initial value of C at t=0

r is the decaying rate, expressed in decimal

The concentration of the pollutants starts at Co=5 mg/lt. We also know the pollutant reduces its concentration by 10% each hour. This gives us a value of r = 10% / 100 = 0.1

Substituting into the general equation:

$C(t)=5\cdot(1-0.1)^t$

Operating:

$\boxed{C(t)=5\cdot(0.9)^t}$

7 0

2 years ago