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morpeh [17]
3 years ago
8

you want to treat some of your friends to the movies, and you have $75 to spend. If each ticket costs $10, and you want to purch

ase $25 worth of snacks, how many friends can you take to the movies? 
Mathematics
1 answer:
netineya [11]3 years ago
5 0

Answer:

5 friends

Step-by-step explanation:

Since the snacks are worth $25, it is deducted from the total

10x=75-25

10x=50

x=5

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Consider the equations 5x+10=30 and 5(x+10)=30. Do they have the same solution? Why or why not?
hodyreva [135]
If you simplify (take out the brackets) of this equation. 5(x+10)=30 then it would be

5 times x + 5 times 10 = 30

5x+10=30

So yes they have the same solution

3 0
2 years ago
PLEASEEE HELP FAST
Viktor [21]

Answer:

1. 4 as that is when he is closest.

2. 3 is when he is waiting as the graph is horizontal showing no movement

3. 4 would change as that is showing his pace walking home if it increased the slope would become steep but if it decreased it would level out more.

Step-by-step explanation:

4 0
3 years ago
May someone please help me ASAP
Ivenika [448]

Answer:

84+n=342

Step-by-step explanation:

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3 0
3 years ago
National data indicates that​ 35% of households own a desktop computer. In a random sample of 570​ households, 40% owned a deskt
Elan Coil [88]

Answer:

Yes, this provide enough evidence to show a difference in the proportion of households that own a​ desktop.

Step-by-step explanation:

We are given that National data indicates that​ 35% of households own a desktop computer.

In a random sample of 570​ households, 40% owned a desktop computer.

<em><u>Let p = population proportion of households who own a desktop computer</u></em>

SO, Null Hypothesis, H_0 : p = 25%   {means that 35% of households own a desktop computer}

Alternate Hypothesis, H_A : p \neq 25%   {means that % of households who own a desktop computer is different from 35%}

The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;

                                  T.S.  = \frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }  ~ N(0,1)

where, \hat p = sample proportion of 570​ households who owned a desktop computer = 40%

            n = sample of households = 570

So, <u><em>test statistics</em></u>  =  \frac{0.40-0.35}{{\sqrt{\frac{0.40(1-0.40)}{570} } } } }

                               =  2.437

<em>Since, in the question we are not given with the level of significance at which to test out hypothesis so we assume it to be 5%. Now at 5% significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics doesn't lies within the range of critical values of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.</em>

Therefore, we conclude that % of households who own a desktop computer is different from 35%.

3 0
2 years ago
Find the distance, c, between (–3, –1) and (3, 2) on the coordinate plane. Round to the nearest tenth if necessary ??
astra-53 [7]

Point distance formula:

d=√((x_2-x_1)²+(y_2-y_1)²)

Substitute

d=√((3-(-3))²+(2-(-1)²)

Simplify

d=√((6)²+(3)²)

d=√(36+9)

d=√(45)

<h2><u><em> d=3√(5)</em></u></h2>

<u><em></em></u>

-Hunter

8 0
2 years ago
Read 2 more answers
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