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liraira [26]
3 years ago
9

1 1/6 - 1 5/8 =...........?

Mathematics
2 answers:
ruslelena [56]3 years ago
4 0

Answer:

-0.45833333333

Step-by-step explanation:

I made them decimals. :)

Alina [70]3 years ago
4 0

Answer:

1\24

Step-by-step explanation:

You might be interested in
Adult tickets to space amusement parkcost x dollars. Children's tickets cost y dollars. The henson family bought 3 aduly and 1 c
vesna_86 [32]

Answer:

3x + 1y = $163

2x + 3y =$174

Step-by-step explanation:

Adult tickets to space amusement park cost x dollars children’s tickets cost y dollars.

Henson family bought 3 adults and 1 child's tickets for $163

x + y = 163

3x + 1y = 163

The Garcia family bought 2 adults and 3 child’s ticket for $174.

x + y = $174

2x + 3y = $174

Therefore the equations representing both families cost are

Henson family

3x + 1y = $163

Garcia family

2x + 3y = $174

3 0
4 years ago
I NEED HELP PLS I WILL MARK BRAINLIEST
ANEK [815]

Answer:

\approx 11.2\:\mathrm{in}

Step-by-step explanation:

The swing of the tip of pendulum is creating an arc. The question is actually asking for the length of this arc. The length of an arc is given by 2r\pi\cdot \frac{\theta}{360}, where r is the radius of the circle and \theta is the angle of the arc.

In this problem, we're given:

\theta = 40^{\circ},\\r=16

Substituting given values, we get:

2\cdot 16\cdot \pi \cdot \frac{40}{360}=32\pi\cdot \frac{1}{9}\approx \boxed{11.2\:\mathrm{in}}

8 0
3 years ago
Could you help me with this for 50? points.
ruslelena [56]
+1,-1 and your other would be 1,1
3 0
2 years ago
Read 2 more answers
A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.
nika2105 [10]

Answer:

The best point of estimate for the true mean is:

\hat \mu = \bar X = 2.55

2.55-1.96\frac{12}{\sqrt{76}}=-0.148    

2.55+1.96\frac{12}{\sqrt{76}}=5.248    

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

\bar X=2.55 represent the sample mean for the late time for a flight

\mu population mean

\sigma=12 represent the population deviation

n=76 represent the sample size  

Confidence interval

The best point of estimate for the true mean is:

\hat \mu = \bar X = 2.55

The confidence interval for the true mean is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

The Confidence level given is 0.95 or 95%, th significance would be \alpha=0.05 and \alpha/2 =0.025. If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got z_{\alpha/2}=1.96

Replacing we got:

2.55-1.96\frac{12}{\sqrt{76}}=-0.148    

2.55+1.96\frac{12}{\sqrt{76}}=5.248    

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0
3 years ago
Solve: 6/x + 23 = 1/y for y.
iris [78.8K]

Answer:

\frac{y}{x}  =  \frac{6y {}^{2} }{x {}^{2} }

3 0
3 years ago
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