A person can teyp 150 words in 5 minutes at the persons rate how many words can that person tyep in one minite

Sweater that originally cost $40 is on sale for 10 percent off. What is the amount of the discount?

saw5 [17]

multiply 40 by 10%

40*0.10 = 4, so the discount would be $4

8 0

3 years ago

Please help solve algebra question

Ksenya-84 [330]

Let be:Speed of the wind: WSpeed of the plane in still air: P
Against the wind the plane flew:Distance: d=175 milesTime: ta=1 hour 10 minutesta=1 hour (10 minutes)*(1 hour/60 minutes)ta=1 hour + 1/6 hourta=(6+1)/6 hourta=7/6 hourSpeed against the wind: Sa=d/taSa=(175 miles) / (7/6 hour)Sa=175*(6/7) miles/hourSa=1,050/7 miles per hourSa=150 mph
(1) P-W=Sa(1) P-W=150
The return trip only took 50 minutesDistance: d=175 milesTime: tr=50 minutestr=(50 minutes)*(1 hour/60 minutes)tr=5/6 hour
Speed retur trip: Sr=d/trSr=(175 miles) / (5/6 hour)Sr=175*(6/5) miles/hourSr=1,050/5 miles per hourSr=210 mph
(2) P+W=Sr(2) P+W=210
We have a system of 2 equations and 2 unknows:(1) P-W=150(2) P+W=210
Adding the equations:P-W+P+W=150+2102P=360Solving for P:2P/2=360/2P=180
Replacing P by 180 in equation (2):(2) P+W=210180+W=210
Solving for W:180+W-180=210-180W=30
Answers:The speed of the plane in still air was 180 mphThe speed of the wind was 30 mph

8 0

3 years ago

Unit 4 Test, Part 2: Congruence and Constructions<br> Help with my math please

lakkis [162]

Answer/Step-by-sep explanation:

1. Given:

∆NMK ≅ ∆TRP

$\overline{NM} = 20$

$\overline{MK} = 15$

$\overline{KN} = 25$

$\overline{TR} = 3x - 1$

a. To complete the congruent statement, thus: ∆MNK ≅ ∆RTP

b. The side that is congruent to $\overline{TR}$ is $\overline{NM}$ . Thus:

$\overline{TR}$ ≅ $\overline{NM}$

c. Since $\overline{TR}$ ≅ $\overline{NM}$ , therefore:

$\overline{TR} = \overline{NM}$

$3x - 1 = 20$ (substitution)

Add 1 to both sides

$3x = 20 + 1$

$3x = 21$

Divide both sides by 3

$x = \frac{21}{3}$

$x = 7$

2. a. Slope of LK = $\frac{rise}{run} = \frac{4}{3}$

Slope of LM = $\frac{rise}{run} = -\frac{3}{5}$

b. ✍️Length of LK is the distance between L(-7, 4) and (-4, 8):

$Lk = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$Lk = \sqrt{(-4 -(-7))^2 + (8 - 4)^2}$

$Lk = \sqrt{(3)^2 + (4)^2}$

$Lk = \sqrt{9 + 16}$

$Lk = \sqrt{25}$

$Lk = 5$

✍️Length of LM is the distance between L(-7, 4) and (-2, 1):

$LM = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$LM = \sqrt{(-2 -(-7))^2 + (1 - 4)^2}$

$LM = \sqrt{(5)^2 + (-3)^2}$

$LM = \sqrt{25 + 9}$

$LM = \sqrt{34}$

$LM = 5.8$ (nearest tenth)

∆KLM is not an isosceles ∆ because it does not has two equal side lengths. This we can see because LK and LM are not equal.

Therefore, Anthony is incorrect. Am isosceles ∆ has two equal sides.

8 0

3 years ago

Solve the system of linear equations y= 2x -3Y= x + 6

Alja [10]

Answer:

x=9, y=15. (9, 15).

Step-by-step explanation:

y=2x-3

y=x+6

----------

2x-3=x+6

2x-x-3=6

x-3=6

x=6+3

x=9

y=9+6=15

7 0

2 years ago

From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean leng

alexandr402 [8]

Complete Question

From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean length of bass caught in a small lake. A normal distribution was assumed. Using the 90% confidence interval obtain:

a. A point estimate of $\mu$ and its 90% margin of error.