Please help me???? 25 points

How many 3 digit numbers can you make using these numbers: 4,2,9 (numbers can be reapeated)

bogdanovich [222]

Answer:

You can make 27 different 3 digit numbers

Step-by-step explanation:

Since we can use 3 different numbers this means that each of the 3 digits can be one of the 3 numbers (4,2,9). This means that there are 3 different values for the first digit, 3 different values for the second digit, and 3 different values for the third digit, which mean that we can make 3 x 3 x 3 different 3 digit numbers (we took the number of possible values for each digit and multiplied them together.

3 x 3 x 3 = 27

8 0

3 years ago

Mary measured the heights of two different plants every day. Plant A was 1 inch tall when

seropon [69]

The heights of plant A and plant B will be equal on the day 8.

<h3>How to find the heights of the trees?</h3>

The height of the tree can be found as follows:

The equation for plant A can be described as follows:

y = 1 + 0.5x

where

x = number of days

y = height

The equation for plant B can be described as follows:

y = 3 + 0.25x

Therefore, for the height to be equal

1 + 0.5x = 3 + 0.25x

1 - 3 = 0.25x - 0.5x

-2 = -0.25x

x = 2 / 0.25

x = 8 days

learn more on height here: brainly.com/question/16959984

#SPJ1

7 0

1 year ago

Anyone wanna help?please

timama [110]

$\large\underline{\underline{\red{\rm\blue{\longmapsto} Step-by-step\: Explanation:-}}}$

Given to points to us are :-

( 3 , -6 )
( -1 , -8 )

( As these are plotted on graph with yellow dots .)

Now , we can use Distance Formula , which is :-

$\boxed{\pink{\tt Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}}$

Here ,

x1 = 3 .
x2 = -1.
y1 = (-6)
y2 = (-8).

<u>→ Substituting the respective values , </u>

⇒ Distance = √ [ { 3 - (-1)}² + { -6 -(-8)²} ] .

⇒ Distance = √ (3+1)² + (8-6)²

⇒ Distance = √ 4² + 2²

⇒ Distance = √ 16 + 4

⇒ Distance = √20 = √4 × √5

⇒ Distance = 4√5units .

<u>Hence the distance between two points is 4√5u.</u>

3 0

3 years ago

Find the appropriate rejection regions for the large-sample test statistic z in these cases. (Round your answers to two decimal

Usimov [2.4K]

Answer:

a) We have that the significance is given by $\alpha =0.01$ and we know that we have a right tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 1% of the area on the right and 99% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.01,0,1)"

And we got for this case $z_{crit}=2.33$

So then the rejection region would be $z>2.33$

b) We have that the significance is given by $\alpha =0.05$ , $\alpha/2 =0.025$ and we know that we have a two tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 2.5% of the area on the right and 97.5% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.025,0,1)"

And we got for this case $z_{crit}=\pm 1.96$

So then the rejection region would be $z>1.96 \cup z$

Step-by-step explanation:

Part a

We have that the significance is given by $\alpha =0.01$ and we know that we have a right tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 1% of the area on the right and 99% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.01,0,1)"

And we got for this case $z_{crit}=2.33$

So then the rejection region would be $z>2.33$

Part b

We have that the significance is given by $\alpha =0.05$ , $\alpha/2 =0.025$ and we know that we have a two tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 2.5% of the area on the right and 97.5% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.025,0,1)"

And we got for this case $z_{crit}=\pm 1.96$

So then the rejection region would be $z>1.96 \cup z$

7 0

3 years ago

If n erasers have a weight of 80 grams, what is the total weight of 50 erasers?

Black_prince [1.1K]

Answer:

$The\ weight\ of\ the\ 50\ erasers\ be\ \frac{4000}{n}\ grams.$