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WITCHER [35]
3 years ago
7

HELP what is the equation of the lines in slope - intercept form

Mathematics
1 answer:
34kurt3 years ago
3 0

Answer:

y=7x

Step-by-step explanation:

slope intercept form is y=mx+b.

m is the slope (rise over run) and b is your y- intercept. you leave y as it is. if the line is going down, you have a negative slope. hope this helped :)

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The expression d/t can be used to find the average speed of an object that travels a distance d in time t. What is a car's avera
wel

hey buddy here is your answer!

hope it helps you..

the answer is C...

we know that,

speed=distance/time

∴s=145/2.5

s=58 mph

6 0
3 years ago
En tu cuaderno realiza la conversión de los siguientes decimales a fracciones.<br>a) 0.250 :<br>​
kenny6666 [7]

Answer:

Tengo que convertir los siguientes números decimales a fracciones, y también tengo que simplificarlos a su mínima expresión. Aquí un ejemplo ;)

(no pido que me digan las respuestas, si no que me ayuden a entenderla)

0.44 = 44/100 = 22/50 = 11/25

0.62=

0.29=

2.05=

1.28=

0.08=

0.50=

0.375=

0.60=

0.7=

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Question 5
galina1969 [7]

Answer:b 1/2

Step-by-step explanation:

7 0
3 years ago
Which phrase SIgnals a compare and-contrast text structure O as a result betore long for this reason on the other hand​
Shalnov [3]

Answer:

D. On the other hand.

Step-by-step explanation:

As a result, before long, and for this reason would be cause and effect. On the other hand is the only remaining option.

7 0
3 years ago
A statistics teacher started class one day by drawing the names of 10 students out of a hat and asked them to do as many pushups
HACTEHA [7]

Answer:  b. 8.56 to 21.44

Step-by-step explanation:

Let  \mu be the  mean number of pushups that can be done.

As per given , we have

Sample size : n= 10

Degree of freedom = n-1=9

Sample mean : \overline{x}=15

Sample standard deviation : s=9

Significance level : α=1-0.95=0.05

From t- distribution table ,

Critical two -tailed t-value for α=0.05 and df = 9 is

t_{\alpha/2, 9}=t_{0.025,9}=2.2622

Confidence interval for \mu is given by :-

\overline{x}\pm t*\dfrac{s}{\sqrt{n}}

=15\pm (2.2622)\dfrac{9}{\sqrt{10}}\\\\\approx 15\pm 6.44=(15-6.44,\ 15+6.44)\\\\=(8.56,\ 21.44)

Hence, the  95% confidence interval for the true mean number of pushups that can be done is 8.56 to 21.44.

5 0
3 years ago
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