What is the measure of angle M?

Would it be 90 or 50? i’m not sure

astra-53 [7]

I'm pretty sure it would be 50°

3 0

3 years ago

Reduce to simplest form.<br> -7/12 +3/8

faust18 [17]

Answer:

-5/24

Step-by-step explanation:

here u gooo :).....

4 0

3 years ago

For the right triangles below, find the exact values of the side lengths b and d.

Ivanshal [37]

Answer:

Step-by-step explanation:

The question lacks the required diagram. Find the diagram attached below;

According to the first triangle, taking 30° as the reference angle, the opposite side of the triangle will be 5 and the adjacent will be the unknown side "b"

According to SOH, CAH, TOA;

tanθ = opposite/adjacent (using TOA)

Given;

θ = 30°, opposite = 5 and adjacent = b

tan30° = 5/b

b = 5/tan30°

b = 5/(1/√3)

b = 5*√3/1

b = 5√3

According to the 45° triangle, the opposite side of the triangle will be d and the hypotenuse will be 7

Using SOH;

sinθ = opposite/hypotenuse

Given;

θ = 45°, opposite = d and adjacent = 7

sin45° = d/7

d = 7sin45°

d = 7(1/√2)

d = 7/√2

Rationalize 7/√2

= 7/√2*√2/√2

=7√2/2

Hence the value of d is 7√2/2

7 0

3 years ago

The following data comparing wait times at two rides at Disney are listed below: Position Pirates Splash Mountain Sample Size 32

myrzilka [38]

Answer:

a) $(14.68 -18.77) - 2.39 \sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}} =-12.968$

$(14.68 -18.77) + 2.39 \sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}} =4.788$

b) $t=\frac{14.68-18.77}{\sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}}}}=-1.10$

Step-by-step explanation:

Data given and notation

$\bar X_{A}=14.68$ represent the mean for Pirates

$\bar X_{B}=18.77$ represent the mean for Splash Mountain

$s_{A}=11.87$ represent the sample standard deviation for the sample Pirates

$s_{B}=16.79$ represent the sample standard deviation for the sample Slpash Mountain

$n_{A}=32$ sample size selected for Pirates

$n_{B}=30$ sample size selected for Splash Mountain

$\alpha=0.02$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

Part a

The confidence interval would be given by:

$(\bar X_A -\bar X_B) \pm t_{\alpha/2} \sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}$

The degrees of freedom are given by:

$df = n_A +n_B -2 = 32+30-2 = 60$

Since we want 98% of confidence the significance level is $\alpha =1-0.98 =0.02$ and $\alpha/2 =0.01$ , we can find in the t distribution with df =60 a critical value that accumulates 0.01 of the area on each tail and we got:

$t_{\alpha/2}= 2.39$

And replacing we got for the confidence interval:

$(14.68 -18.77) - 2.39 \sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}} =-12.968$

$(14.68 -18.77) + 2.39 \sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}} =4.788$

Part b

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the means are equal, the system of hypothesis would be:

Null hypothesis: $\mu_{A} = \mu_{B}$

Alternative hypothesis: $\mu_{A} \neq \mu_{B}$

the statistic is given by:

$t=\frac{\bar X_{A}-\bar X_{B}}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{14.68-18.77}{\sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}}}}=-1.10$

6 0

3 years ago

A candy store called "Sugar" built a giant hollow sugar cube out of wood to hang above the entrance to their store. It took 5454

Elina [12.6K]

Answer:

27 ft^3

Step-by-step explanation:

The total surface area of the cube is 54 ft^2.

A cube has 6 congruent square sides.

54 ft^2/5 = 9 ft^2

Each square side has area 9 ft^2.

The area of a square is the square of the side, so the side of a square is the square root of the area.

side = sqrt(9 ft)^2 = 3 ft

The volume of a cube is the cube of the side.

volume = (3 ft)^3

volume = 27 ft^3

4 0

3 years ago