Can someone plz help me on this

WILL GIVE BRAINIEST TO VERY FIRST ANSWER

adell [148]

<span>l 18+7 l / 10 - l -16+26 l /6
= 25 / 10 - 10/6
= 5/2 - 5/3
= 15/6 - 10/6
= 5/6

answer </span><span>5/6 (3rd choice)</span>

8 0

3 years ago

Read 2 more answers

The University of Ballygobackwards has three colleges: the College of Science, which has 12 staff, 8 of whom are female; the Col

IceJOKER [234]

Answer:

just split the awnser

Step-by-step explanation:

6 0

3 years ago

What is the mass of a copper (II) chloride sample if it contains 0.0344 moles of the compound?

ale4655 [162]

According to the valence number of copper(2+) and the same value for chlorine (1-) copper (II) chloride has the formula of CuCl2

The molar mass of copper is 0,0635 kg/mole and chlorine gas a molar mass of 0,035 kg/mole the compound will have a molar mass of ( 0,0635+2×0,035 )kg/mole=0,099kg/mole and 0,344 moles are equivalent in mass to 0,344×0,135 kg=0,046 kg

5 0

2 years ago

Suppose that the amount of algae in pound doubles every hour. If the pond initially contains 50 pounds of algae how much algae w

musickatia [10]

Answer:

250

Step-by-step explanation:

because 50 + (4 x 50) that equals 50 + 200 = 250

3 0

3 years ago

Read 2 more answers

preliminary sample of 100 labourers was selected from a population of 5000 labourers by simple random sampling. It was found tha

VladimirAG [237]

Answer:

$n=\frac{0.4(1-0.4)}{(\frac{0.05}{1.96})^2}=368.79$

n=369

Step-by-step explanation:

1) Notation and definitions

$X=40$ number of the selected labourers opt for a new incentive scheme.

$n=100$ random sample taken

$\hat p=\frac{40}{100}=0.4$ estimated proportion of the selected labourers opt for a new incentive scheme.

$p$ true population proportion of the selected labourers opt for a new incentive scheme.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

2) Solution tot he problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by: