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3 years ago

What is the product of 0.4 and 2.5

Mathematics

2 answers:

ser-zykov [4K]3 years ago

7 0

The product of 0.4 x 2.5 equals to 1.

Maksim231197 [3]3 years ago

4 0

.4 × 2.5 = 1. product means multiply / sum would be addition

You might be interested in

In △EKL, m∠K = 90º, m∠E = 25º, EK = 3 cm, KH - altitude. Find EH.

laiz [17]

Answer:

EH = 3.31

Step-by-step explanation:

We have been given a right angle triangle EKL. As KH has been given as the altitude (perpendicular) of the right angled triangle, and K is the right angle, we can say that EK is tthe base of the triangle and EH is the only side lleft, which is the hypotenuse of the triangle.

Where,

EK = Base = 3

KH = perpendicular altitude

EH = Hypotenuse

m<K = 90

m<E = 25

We know that

cosθ = Base/ Hypotenuse

cos 25 = 3/ EH

EH = 3/cos25

EH = 3.31

Perpendicular alitutude can also be calculated by using the formula for tanθ.

3 0

3 years ago

Here's another one :)

jekas [21]

Area Of Triangle:

4/2 = 2
2x6 = 12

12x2 = 24ft^2

Remaining area:

150 - 24 = 126ft^2

Maximum length:

126/6 = 21

x = 21ft

8 0

3 years ago

Please help fast Find the volume of a right circular cone that has a height of 15-3 ft and a base with a

mars1129 [50]

Answer:

3095.6 ft³

Step-by-step explanation:

Volume cone = 1/3(πr²h)

= 1/3(π(13.9)²(15.3))

= 1/3(π(193.21)(15.3))

= 1/3(2956.113π)

= 985.371π ≈ 3095.6 ft³

-Chetan K

4 0

3 years ago

Classify each pair of labeled angles as complementary, supplementary, or neither.

stiv31 [10]

Answer:

Step-by-step explanation:

A) neither since the angles add to 179 deg.

B) neither since 63 + 47 = 110.

C) complementary since 61 + 29 = 90

3 0

2 years ago

g An irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the

blagie [28]

Answer:

A 90% confidence interval of the true mean is [$119.86, $123.34].

Step-by-step explanation:

We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.

Also, the standard deviation of the population was $6.36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean amount of money spent on textbooks = $121.60

$\sigma$ = population standard deviation = $6.36

n = sample of students = 36

$\mu$ = population mean

Here for constructing a 90% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $121.60-1.645 \times {\frac{6.36}{\sqrt{36} } }$ , $121.60+1.645 \times {\frac{6.36}{\sqrt{36} } }$ ]

= [$119.86, $123.34]

Therefore, a 90% confidence interval of the true mean is [$119.86, $123.34].

5 0

3 years ago