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2 years ago

13

Slope of 3 and passes through (-2,-9) please helpppp. :(

1 answer:

Andrei [34K]2 years ago

3 0

Answer: y=-3x-2

Explanation:
Given;

Slope of the line -3

Point (-2,4)

Use the formula: mx+c=y

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Denise is making bean bags. She has 6.4 pounds of beans.

dezoksy [38]

Answer:

a) she be able to make 8 bags

b) she can make 16 bags.

Step-by-step explanation:

According to the exercise info Denise has 6.4 pounds of beans

<u>Solving question a </u>

⇒ if she makes each bean bag of 0.8 pounds and she has 6.4 pounds of beans, we can express this by using the three simply rule she needs to puts 6.4 pounds of beans in x quantity of bags of 0.8 pouns, so doing the division: 6. 4 pounds of beans / 0.8 pounds of beans per bag= 8 bags of 0.8 pounds of beans.

B- If she decides to make mini bean bags that are half as heave as the ohers bags, so now Denise is going to make bags of 0.8/2 = 0.4 pounds.

Doing the same procedure as question a, we can say that she has 6.4 pounds of beans and she needed to put these amount in a x quantity of 04 pounds bags. Doing the following division ⇒ 6.4 pounds of beans/ 0.4 pounds of beans per bag= 16 bags

In conclusion she is going to make 16 bags of 0.4 pounds of beans.

5 0

3 years ago

Help me plsssssssssssssssssssssssssss

Feliz [49]

Answer:

#1 is 3/6 or 1/2

#3 is 1/6

#5 is 4/6 or 2/3

4 0

3 years ago

Read 2 more answers

43 is a number between

MAXImum [283]

Answer:

42 and 44 is the anwser

Step-by-step explanation:

41 42 43 44 that is the explanation

8 0

2 years ago

Read 2 more answers

17. ralph is 3 times as old as sarah. in 4 years ralph will only be twice as old as sara will be then. find ralph’s age now.

spin [16.1K]

Answer:

Sara is now 4, will be 8 in 4 years. Ralph is now 12, will be 16 in 4 years.

7 0

2 years ago

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

Ivanshal [37]

Answer:

a) $\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

b) $z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

8 0

3 years ago