Use the net to compute the surface area of the three-dimensional figure.

Q2) The number of people signed up for a bus trip increased from 32 to

Oduvanchick [21]

Answer:

45-32 = 13 / 32 *100= 40.625 round to the nearest percent is 41%

Step-by-step explanation: Brainlest plzzz

5 0

3 years ago

The manager went over the sales of mobile phones at the store and found that the mean sale was 45, with a standard deviation of

d1i1m1o1n [39]

The z-score of the sale of mobile phones on that day is 1.75

Step-by-step explanation:

The formula of z-score is z = (x - μ)/σ, where:

x is the score
μ is the mean
σ is the standard deviation

∵ The mean sale was 45

∴ μ = 45

∵ The standard deviation was 4

∴ σ = 4

∵ 52 mobile phones were sold on a particular day

∴ x = 52

To find z-score of the sale of mobile phones on that day substitute the values of x, μ, and σ in the formula of z-score

∵ $z=\frac{52-45}{4}$

∴ $z=\frac{7}{4}$

∴ z = 1.75

The z-score of the sale of mobile phones on that day is 1.75

Learn more:

You can learn more about z-score in brainly.com/question/7207785

#LearnwithBrainly

3 0

3 years ago

a rectangular plot of land 8m long and 6m wide.there is a flower bed in the form of a circle of radius 3m.find 1.area of land. 2

mixer [17]

Answer:

1)Area of the Land is 48 square meters

2) Area of flower bed is 28.26 square meter

3)Area of land excluding flower bed is 19.74 square meters

4)cost of planting grass in land excluding flower bed is Rs 118.44

Step-by-step explanation:

Given:

Length of the Rectangular plot = 8m

Width of the rectangular plot = 6m

Radius of the circular flower bed = 3m

1)Area of the Land

Area of the Rectangular plot = Area of the Land

Area of the Land = length x width

Area of the Land = 8*6

Area of the Land = 48 square meters-----------------------------(1)

2. Area of flower bed

Area of flower bed = 4 x area of one quadrant

Area of flower bed = area of the circle

Area of flower bed = $\pi r^2$

where r is the radius

Now , substituting the values

Area of flower bed = $\pi (3)(3)$

Area of flower bed = 28.26 square meters-----------------------------(2)

3)Area of land excluding flower bed

Area of land excluding flower bed = Area of the rectangular land - area of flower beds

Area of land excluding flower bed = 48 - 28.26

Area of land excluding flower bed = 19.74 square meters

4)cost of planting grass in land

Cost of planting grass per square meter = 6

Excluding the flower bed we have 19.74 square meters. to plant grass for this 19. 74 square meters, the cost will be

= $19. 74 \times 6$

= Rs 118.44

6 0

3 years ago

Tell whether this statement is sometimes, always, or never true. Identify the answer with a sketch. If B is a point between A an

Ksivusya [100]

The statement is always true

7 0

3 years ago

Consider the hypotheses below. Upper H 0: mu equals 50 Upper H 1: mu not equals 50 Given that x overbar equals 60, s equals 1

Neko [114]

Answer:

We conclude that $\mu\neq$ 50.

Step-by-step explanation:

We are given that x bar = 60, s = 12, n = 30, and alpha = 0.10

Also, Null Hypothesis, $H_0$ : $\mu$ = 50

Alternate Hypothesis, $H_1$ : $\mu$ $\neq$ 50

The test statistics we will use here is;

T.S. = $\frac{Xbar-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, X bar = sample mean = 60

s = sample standard deviation = 12

n = sample size = 30

So, Test statistics = $\frac{60-50}{\frac{12}{\sqrt{30} } }$ ~ $t_2_9$

= 4.564

At 10% significance level, t table gives critical value of 1.311 at 29 degree of freedom. Since our test statistics is more than the critical value as 4.564 > 1.311 so we have sufficient evidence to reject null hypothesis as our test statistics will fall in the rejection region.

P-value is given by, P( $t_2_9$ > 4.564) = less than 0.05% {using t table}

Here also, P-value is less than the significance level as 0.05% < 10% , so we will reject null hypothesis.

Therefore, we conclude that $\mu\neq$ 50.

7 0

3 years ago