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Andre45 [30]
3 years ago
12

Please help I can't figure this out!!!!!

Mathematics
1 answer:
Kamila [148]3 years ago
8 0
M < 2 = 1/2 * 120   = 60 
m < 3  = 90 - 60 = 30 degrees
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Jack is a car salesman who works strictly on commission. For every car he sells, he earns a 2% commission. Cars sell for $35,000
Nadya [2.5K]
First we need to find 2% of 35,000

2% equals 0.02 so we'll multiply 35,000 by 0.02

35,000 × 0.02 = 700

for every car he sells jack gets $700

now we need to divide that into 5,000

5,000 ÷ 700 = 7.14

now since you cannot sell 0.14 of a car you need to round up

he needs to sell 8 cars to meet his target income
4 0
3 years ago
I don’t understand this please help
Mrrafil [7]

Answer:

i think BC = 8.5

Step-by-step explanation:

19-(2+6.5+2)

19-10.5

8.5

7 0
3 years ago
Find the slope of the line passing through each of the following pairs of points. a (−7, −4), (−6, −8)
Romashka [77]

Answer:

m= --4

Step-by-step explanation:

Slope (m) = ΔY /ΔX  = (-8-(-4))/(-6-(-7) =  -4/1  = -4

7 0
4 years ago
Read 2 more answers
A phone manufacturer wants to compete in the touch screen phone market. Management understands that the leading product has a le
CaHeK987 [17]

Answer:

Step-by-step explanation:

Hello!

To compete in the touch screen phone market a manufacturer aims to release a new touch screen with a battery life said to last more than two hours longer than the leading product which is the desired feature in phones.

To test this claim two samples were taken:

Sample 1

X: battery lifespan of a unit of the new product (min)

n= 93 units of the new product

mean battery life X[bar]= 8:53hs= 533min

S= 84 min

Sample 2

X: battery lifespan of a unit of the leading product (min)

n= 102 units of the leading product

mean battery life X[bar]= 5:40 hs = 340min

S= 93 min

The population variances of both variances are unknown and distinct.

To test if the average battery life of the new product is greater than the average battery life of the leading product by 2 hs (or 120 min) the parameters of interest will be the two population means and we will test their difference, the hypotheses are:

H₀: μ₁ - μ₂ ≤ 120

H₁:  μ₁ - μ₂ > 120

Considering that there is not enough information about the distribution of both variables, but both samples are big enough, we can apply the central limit theorem and approximate the distribution of both sample means to normal, this way we can use the standard normal:

Z= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2}  } }

Z≈N(0;1)

Z= \frac{(533-340)-120}{\sqrt{\frac{84^2}{56} +\frac{93^2}{102}  } }= 5.028

I hope this helps!

3 0
4 years ago
Someone please help me with the answer. ASAP
labwork [276]

Answer:

Option D is correct

Step-by-step explanation:

-4(3x/2 - 1/2) = -15

<=>

-4*3x/2 + (-4)*(-1/2) = -15

<=>

-6x + 2 = -15

4 0
3 years ago
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