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schepotkina [342]

3 years ago

15

a boat travels 45 minutes at a speed of 8 mph. then the boat travels 1.5 times faster for 1/3 hr. how many miles do the boat tra

vel in all. need help ASAP!!!

1 answer:

Serga [27]3 years ago

7 0

Answer:

10 miles

Step-by-step explanation:

45/60 x 8 = 6 miles

8 x 1.5 = 4 miles

Now, add the two products together.

6 + 4 = 10 miles.

The boat traveled 10 miles in total.

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(25+3) - 7 x 3 + (2x4)

arlik [135]

Answer:

It is not correct, the answer is 15

Step-by-step explanation:

They forgot the rule of PEMDAS, and accidentally subtracted before they multiplied. It should've gone

(25+3) - 7 x 3 + (2x4)

28-7x3+8

28-21+8

Since it's minus 21, that means it's negative, so -21+8=13

28-13=15

4 0

3 years ago

Which of these is a simplified form of the equation 7y + 8 = 9 + 3y + 2y? 7y = 6 5y = 17 12y = 17 2y = 1

LekaFEV [45]

7y +8= 9+3y+2y

Combine like terms.

7y+8= 9+5y

Subtract 5y on both sides.

2y+8=9

Subtract 8 on both sides.

2y=1

Your answer is 2y=1

I hope this helps!
~kaikers

4 0

3 years ago

Read 2 more answers

Plz help ASAP! 10 points

loris [4]

There’s this app called Socratic that you can use. It’s like this one but it’s a little different

4 0

2 years ago

What is the length of BD?

Mademuasel [1]

Answer:

From -6 to positive 1 the length is 7

Step-by-step explanation:

you still add'em.

7 0

2 years ago

Read 2 more answers

Given the following observed and expected data (total of 1000), using chi-squared calculate a p-value that corresponds with this

finlep [7]

Answer:

The answer is " $0.90>p>0.75.$ "

Step-by-step explanation:

$\text{Cinnabar vestigial} \ \ \ \ \ \ \ \ \ \ \ 384 \ \ \ \ \ \ \ \ \ \ \ 390 \ \ \ \ \ \ \ \ \ \ \ -6 36 \ \ \ \ \ \ \ \ \ \ \ 0.092308\\\\$

$roof \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 408 \ \ \ \ \ \ \ \ \ \ \ 390 \ \ \ \ \ \ \ \ \ \ \ 18 \ \ \ \ \ \ \ \ \ \ \ 324 \ \ \ \ \ \ \ \ \ \ \ 0.830769\\\\\text{Cinnabar vestigial roof} \ \ \ \ \ \ \ \ \ \ \ \ \ 63 \ \ \ \ \ \ \ \ \ \ \ \ \ 70\ \ \ \ \ \ \ \ \ \ \ \ -7 \ \ \ \ \ \ \ \ \ \ \ 49 \ \ \ \ \ \ \ \ \ \ \ 0.7\\\\\text{wild type} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 72 \ \ \ \ \ \ \ \ \ \ \ 70 \ \ \ \ \ \ \ \ \ \ \ 2 \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ 0.057143\\\\$

$vestigial \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 32 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 35 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.257143\\\\$

$\text{Cinnabar roof} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 34\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 35 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.028571\\\\\text{roof vestigial} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ -1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.2\\\\$

$\text{cinnabar} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.8 \\\\$

$Total \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1000 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.965934$

Eight phenotypes were present.

Df is provided also by a number of phenotypes -1 The degree of freedom

$\to df = 8-1= 7$

For p-value 0,9, Chi-square is 2.83;

The p-value of 0.75 is 4.5. Chi-square

Chi-sqaure value is observed at 2.965.

That means 0.90>p-value>0.75.

7 0

3 years ago