Step-by-step explanation:
Follow my steps. Sorry im bit untidy .
Follow these 1°,2°,3°
Complete Question
Determine whether the normal sampling distribution can be used. The claim is p < 0.015 and the sample size is n=150
Answer:
Normal sampling distribution can not be used
Step-by-step explanation:
From the question we are told that
The null hypothesis is 
The alternative hypothesis is 
The sample size is n= 150
Generally in order to use normal sampling distribution
The value 
So


Given that
normal sampling distribution can not be used
Hey there!
<u>Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0</u>
x = -4 or x = 2 ✅
<em><u>Quadratic</u></em><em><u> </u></em><em><u>formula </u></em><em><u>:</u></em><em><u> </u></em>ax² + bx + c = 0 where a ≠ 0
The number of real-number solutions <em>(roots)</em> is determined by the discriminant (b² - 4ac) :
- If b² - 4ac > 0 , There are 2 real-number solutions
- If b² - 4ac = 0 , There is 1 real-number solution.
- If b² - 4ac < 0 , There is no real-number solution.
The <em><u>roots</u></em> of the equation are determined by the following calculation:

Here, we have :
1) <u>Calculate </u><u>the </u><u>discrim</u><u>i</u><u>n</u><u>ant</u><u> </u><u>:</u>
b² - 4ac ⇔ 2² - 4(1)(-8) ⇔ 4 - (-32) ⇔ 36
b² - 4ac = 36 > 0 ; The equation admits two real-number solutions
2) <u>Calculate </u><u>the </u><u>roots </u><u>of </u><u>the </u><u>equation</u><u>:</u>
▪️ (1)

▪️ (2)

>> Therefore, your answers are x = -4 or x = 2.
Learn more about <u>quadratic equations</u>:
brainly.com/question/27638369
Step-by-step explanation:Let number of 49 cents =A
Let number of 33 cents=B
<u>A+B=45 equation 1</u>
1 $=100 cents
49 cents =.49 dollar
33 cents=.33 dollar
so .<u>49A+.33B=17.89 equation 2</u>
Add equation 1 and equation 2
since these are simultaneous equations
.49A+.33B=17.89
A+ B=45
so divide equation 2 by 100 we get
49A+33B=1789
A+B =45 multiply equation 1 by 33 on both the sides we get
<u>33A+33B=1485 equation 3</u>
so subtract equation 3 from equation 2 we get
49A+33B=1789 equation 2
<u>33A+33B=1485 equation 3</u>
<u>16A+0 =304</u>
16A=304
A=304/16
A=19
putting nalue of A in equation 1 we get
A+B=45
19+B=45
B=45-19
B=26
so number of 49 cent stamps are 19
and number of 33 cent stamps are 26