All categories

3 years ago

2w=56 what is in simplest form

Mathematics

1 answer:

Ivanshal [37]3 years ago

6 0

2w= 56
/2 /2
------------
w= 28

You might be interested in

Rick knows that 1 cup of glue weighs 1/18 pound. he has 2/3 pound of glue. how many cups of glue does he have

Mumz [18]

(2/3) / (1/18) =
2/3 * 18/1 =
36/3 =
12 cups <===

4 0

3 years ago

Read 2 more answers

A bag contains 2 red, 4 blue, and 4 yellow marbles. What is the probability of pulling a red marble, putting the marble back, an

sdas [7]

Answer:

B.) 8%

Step-by-step explanation:

Answer: B.) 8%

8 0

3 years ago

Find the first four terms of a sequence using the recursive definition. Please help

Ymorist [56]

Answer:

2, 5, 9, 19

Step-by-step explanation:

Using the recursive formula f(n) = 2f(n - 2) + f(n - 1)

with f(1) = 2 and f(2) = 5

f(3) = 2f(3 - 2) + f(3 - 1)

= 2f(1) + f(2) = (2 × 2) + 5 = 4 + 5 = 9

f(4) = 2f(4 - 2) + f(4 - 1)

= 2f(2) + f(3) = (2 × 5) + 9 = 10 + 9 = 19

7 0

3 years ago

Can someone help please? this is due soon. ill give brainliest

Iteru [2.4K]

Not sure but I think it’s B

4 0

2 years ago

Read 2 more answers

Jacob Lee is a frequent traveler between Los Angeles and San Diego. For the past month, he wrote down the flight times in minute

valkas [14]

Solution :

We know that

$H_0: \mu_1 = \mu_2=\mu_3$

$H_1 :$ At least one mean is different form the others (claim)

We need to find the critical values.

We know k = 3 , N = 35, α = 0.05

d.f.N = k - 1

= 3 - 1 = 2

d.f.D = N - k

= 35 - 3 = 32

SO the critical value is 3.295

The mean and the variance of each sample :

Goust Jet red Cloudtran

$\overline X_1 =50.5$ $\overline X_2 =50.07143$ $\overline X_3 =55.71429$

$s_1^2=19.96154$ $s_2^2=14.68681$ $s_3^2=36.57143$

The grand mean or the overall mean is(GM) :

$\overline X_{GM}=\frac{\sum \overline X}{N}$

$=\frac{51+51+...+49+49}{35}$

= 52.1714

The variance between the groups

$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$

$=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$

$=\frac{127.1143}{2}$

= 63.55714

The Variance within the groups

$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$

$=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$

$=\frac{669.8571}{32}$

= 20.93304

The F-test statistics value is :

$F=\frac{s_B^2}{s_W^2}$

$=\frac{63.55714}{20.93304}$

= 3.036212

Now since the 3.036 < 3.295, we do not reject the null hypothesis.

So there is no sufficient evidence to support the claim that there is a difference among the means.

The ANOVA table is :

Source Sum of squares d.f Mean square F

Between 127.1143 2 63.55714 3.036212

Within 669.8571 32 20.93304

Total 796.9714 34

3 0

2 years ago