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belka [17]
3 years ago
13

Amna has 50 points in her shopping card (AED 5 = 1 Point). She spends AED 85 a week for groceries, how many points will be there

in her shopping card after 4 weeks.
Mathematics
1 answer:
tia_tia [17]3 years ago
4 0
Well I’m pretty sure that if 5=1 then 85 a week times 4 weeks is 340, so I guess there would be 340 points after 4 weeks?
Sorry if I’m wrong, I’m not the greatest at math :/
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MrRissso [65]

See the picture above for the solution.

4 0
3 years ago
What is the solution of the following system of equations?<br> 2x + y = -7<br> -3x - 2y = 10
pentagon [3]

Answer:

the solution is    (-4, -3)

Step-by-step explanation:

Multiply the first equation by 2.  This produces 4x + 2y = -14.

Now combine this result with the second equation:

4x + 2y = -14

-3x - 2y = 10

------------------

 x          = -4

Subbing -4 for x in the first equation yields 2x - 4 = -7, or 2x = -3.  

Then x = -3/2, and the solution is    (-4, -3)

8 0
2 years ago
Brainliest
NemiM [27]

Answer:

Infinitely Many

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3 0
3 years ago
Read 2 more answers
Test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students at the 0.005 signif
algol [13]

Answer:

a) F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

b) B. two-tailed

d) t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64  

e) p_v =P(t_{118}  

f) B. Reject the null hypothesis

Step-by-step explanation:

Information provided

\bar X_{O}=2.91 represent the mean for the Orange Coast

\bar X_{C}=2.96 represent the mean for the Coastline

s_{O}=0.05 represent the sample standard deviation for Orange Coast

s_{C}=0.03 represent the sample standard deviation for Coastline

n_{O}=60 sample size for Orange Coast

n_{C}=60 sample size for Coastline

\alpha=0.005 Significance level provided

t would represent the statistic

Part a

For this case we want to test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students

Null hypothesis:\mu_{O} \geq \mu_{C}  

Alternative hypothesis:\mu_{O} < \mu_{C}  

F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

Part b

For this case we need to conduct a left tailed test.

B. two-tailed

Part d

The statistic is given by:

t=\frac{(\bar X_{O}-\bar X_{C})-\Delta}{\sqrt{\frac{s^2_{O}}{n_{O}}+\frac{s^2_{C}}{n_{C}}}} (1)  

And the degrees of freedom are given by df=n_1 +n_2 -2=60+60-2=118  

Replacing the info we got:

t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64  

Part e

We can calculate the p value with this probability:

p_v =P(t_{118}  

Part f

Since the p value is a very low value compared to the significance level given of 0.005 we can reject the null hypothesis.

B. Reject the null hypothesis

5 0
2 years ago
If 5x:(x-2) = 11:3 , find the value of x ?​
Masteriza [31]

Answer:

x = -11/2

Step-by-step explanation:

The ratios need to stay the same

5x        11

----- = -------

x-2       3

Using cross products

5x*3 = (x-2) 11

15x = 11x - 22

Subtract 11 from each side

15x-11x = -22

4x = -22

x = -22/4

x = -11/2

5 0
2 years ago
Read 2 more answers
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