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3 years ago

5 language books, 6 math books. in how many ways can one math book and one language book be paired?

1 answer:

5 0

Subtract 6 and 5. The result will be 1, which will remove 1 book to which has more numbers. 6 Math books, you take one and they are 55 language books5 books of mathematics.If in the problem tells you that you should not remove any book ... well idk lol

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Help meeeeeeeeeeeeeeeweeeeeeeeeeeeeeeee

Yuri [45]

Step-by-step explanation:

This can be put into fractions like this

Say x is the length of the missing dimension

8.5/11 = 10.5/x

Because the 2 drawings are SIMILAR, the ratio of their side lengths has to also be the same

8.5 ÷ 10.5 = 0.8095

11 × 0.8095 = 8.9047

Roung the answer to the nearest tenth

x = 8.9 in

5 0

3 years ago

River C is 400 miles longer than River D. If the sum of their lengths is 5 comma 520 miles, what is the length of each river?

Viefleur [7K]

If the sum of their lengths is 5,520 miles then River D is 2360 and River C is 3160

4 0

3 years ago

What is the vertex of the parabola shown?

Ivenika [448]

10,800 would be the best awnser for you

3 0

3 years ago

Read 2 more answers

PLEASE HELP :((((

sertanlavr [38]

88(3.7+4.3)-2*2.6

88(8)-2*2.6

704-2*2.6

704-5.2

698.8

I hope it helps!

Please mark Brainliest if it does!:-)

5 0

2 years ago

A researcher collects data on the number of times race horses are raced during their careers. The veterinarian finds that the av

vesna_86 [32]

Answer:

95% confidence for µ, the average number of times a horse races

(12.493 , 18.107)

Step-by-step explanation:

Explanation:-

The veterinarian finds that the average number of races a horse enters is 15.3

The mean of the sample x⁻ = 15.3

Given standard deviation of the sample 'S' = 6.8

Given sample size 'n' = 25

Degrees of freedom = n-1 =25-1 =24

The tabulated value 't' = 2.064 at two tailed test 0.95 level of significance

95% confidence for µ, the average number of times a horse races

$(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } , x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )$

$(15.3- 2.064 \frac{6.8}{\sqrt{25} } , 15.3 + 2.064\frac{6.8}{\sqrt{25} } )$

(15.3 - 2.80704 ,15.3 +2.80704)

(12.493 , 18.107)

Conclusion:-

95% confidence for µ, the average number of times a horse races

(12.493 , 18.107)

3 0

3 years ago