M<6 is (2x - 5)º and m<8 is (x + 5)º.<br> What is m<3?<br> 8

Evaluate each function.<br> 1) f(x) = x2 + 4x; Find f(-7)

katrin [286]

Answer:

The answer is -77

Step-by-step explanation:

Ok, so assuming by x2 you mean x squared, I will solve this. So basically when you have a function, f(-7) would mean that you would have to replace all the x's in the equation with -7. So let's write that out. that would be f(-7) = -7^2 + (-7*4). So now according to PEMDAS, you would solve the exponent first, and -7^2 is equal to -49, because when you solve it you would do -(7^2), which is -(49), which is then -49. So now you have f(-7)= -49+ (4*-7). Solving for (4*-7), you get -28. This leaves you with -49 + (-28), which is -49 - 28. Simplifying that, you get the answer, which is -77.

5 0

3 years ago

Fill in the missing blinks

Natalka [10]

3-4: Definition of supplementary angles
5: Simplify

4 0

3 years ago

Carlos has 38 apples.He puts 8 into each bag.If he fills many bags as possible ,how many apples are left?

leva [86]

Well that would be 8 apples in 4 bags so there’s 6

answer: 6

7 0

3 years ago

Witch is grater 18/24 or 55%

GrogVix [38]

Answer:

i would say 18/24

Step-by-step explanation:

6 0

3 years ago

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: