Confused PLEZZZZZ HELLLPPPP

How do I know which operation to use with a word math problem

Mamont248 [21]

Answer:

you use your biggest brain cell

Step-by-step explanation:

4 0

3 years ago

(05.06) A bicycle manufacturing company makes a particular type of bike. Each child bike requires 4 hours to build and 4 hours t

Anestetic [448]

Answer: The company can build 10 child bikes and 12 adult bikes in the week .

Step-by-step explanation:

Let 'c' be the number of child bike and 'a' be the number of adult bike.

According to the problem

The restriction of building time for a week is 4c+6a≤120 hours.........(1)

and the restriction of testing time for a week is 4c+4a≤100 hours...............(2)

Lets check whether company can build c=10 and a=12 bikes in a week by putting this value in (1) and (2)

(1)......4(10)+6(12)=40+72=112≤120 ⇒Restriction of building time is satisfied.

(2)......4(10)+4(12)=40+48=88≤100⇒Restriction of testing time is satisfied.

Hence, the company can build 10 child bikes and 12 adult bikes in the week,as order is meeting with the restrictions.

4 0

3 years ago

Read 2 more answers

Find all real zeros of the function y=2x-4

Ber [7]

For this case we have the following linear function:
$y = 2x-4
$
What we must do is equal the function to zero or equivalently, substitute the value of y = 0 in the given function.
We have then:
$2x - 4 = 0
$
From here, we clear the value of x.
We have then:
$2x = 4

x = 4/2

x = 2$
Answer:
all real zeros of the function y = 2x-4 are:
x = 2

5 0

3 years ago

10) Megan's started a service club at her high school. After all of the members signed up

bogdanovich [222]

Answer:

Your answer is 55

Step-by-step explanation:

11% of 500 is 55

hope it helps :)

pls mark brainliest :P

8 0

3 years ago

A simple random sample of 60 households in city 1 is taken. In the sample, there are 45 households that decorate their houses wi

murzikaleks [220]

Answer:

The calculated value of z = - 0.197 falls in the critical region therefore we reject the null hypothesis and conclude that at the 5% significance level there is significant difference in population proportions of households that decorate their houses with lights for the holidays

Step-by-step explanation:

We formulate the null and alternative hypotheses as

H0: p1= p2 there is no difference in population proportions of households that decorate their houses with lights for the holidays

against Ha : p1≠ p2 (claim) ( two sided)

The significance level is set at ∝= 0.05

The critical value for two tailed test at alpha=0.05 is ± 1.96

or Z∝= 0.05/2= ± 1.96

The test statistic is

Z = p1-p2/√pq(1/n1 +1/n2)

p1= proportions of households decorating in city 1 = 45/60=0.75

p2= proportions of households decorating in city 2 = 40/50= 0.8

p = the common proportion on the assumption that the two proportion are same.

p = $\frac{n_1p_1 +n_2p_2}{n_1+n_2}$

Calculating

p =60 (0.75) + 50 (0.8) / 110

p= 45+ 40/110= 85/110 = 0.772

so q = 1-p= 1- 0.772= 0.227

Putting the values in the test statistic and calculating

z= 0.75- 0.8/ √0.772*0.227( 1/60 + 1/50)

z= -0.05/√ 0.175244 ( 110/300)

z= -0.05/0.25348

z= -0.197

The calculated value of z = - 0.197 falls in the critical region therefore we reject the null hypothesis and conclude that at the 5% significance level there is significant difference in population proportions of households that decorate their houses with lights for the holidays

4 0

3 years ago